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Osmosis and Osmotic pressure of the solution Questions in English
Class 12 Chemistry · Solutions · Osmosis and Osmotic pressure of the solution
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301
MediumMCQ
The osmotic pressure (in $atm$) of an aqueous solution containing $0.01 \ mol$ of $NaCl$ (degree of dissociation $0.94$) and $0.03 \ mol$ of glucose in $500 \ mL$ at $27^{\circ} C$ is $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$
A
$2.43$
B
$4.23$
C
$3.24$
D
$3.42$
Solution
(A) The osmotic pressure $\pi$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the total molarity,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
First,calculate the van't Hoff factor for $NaCl$: $i = 1 + \alpha(n-1)$. Here $\alpha = 0.94$ and $n = 2$,so $i = 1 + 0.94(2-1) = 1.94$.
The effective moles of $NaCl$ particles = $i \times \text{moles} = 1.94 \times 0.01 = 0.0194 \ mol$.
Glucose is a non-electrolyte,so its moles remain $0.03 \ mol$.
Total moles of solute = $0.0194 + 0.03 = 0.0494 \ mol$.
Volume of solution = $500 \ mL = 0.5 \ L$.
Total molarity $C = \frac{0.0494 \ mol}{0.5 \ L} = 0.0988 \ M$.
Temperature $T = 27 + 273 = 300 \ K$.
Osmotic pressure $\pi = 0.0988 \times 0.082 \times 300 = 2.43048 \ atm \approx 2.43 \ atm$.
First,calculate the van't Hoff factor for $NaCl$: $i = 1 + \alpha(n-1)$. Here $\alpha = 0.94$ and $n = 2$,so $i = 1 + 0.94(2-1) = 1.94$.
The effective moles of $NaCl$ particles = $i \times \text{moles} = 1.94 \times 0.01 = 0.0194 \ mol$.
Glucose is a non-electrolyte,so its moles remain $0.03 \ mol$.
Total moles of solute = $0.0194 + 0.03 = 0.0494 \ mol$.
Volume of solution = $500 \ mL = 0.5 \ L$.
Total molarity $C = \frac{0.0494 \ mol}{0.5 \ L} = 0.0988 \ M$.
Temperature $T = 27 + 273 = 300 \ K$.
Osmotic pressure $\pi = 0.0988 \times 0.082 \times 300 = 2.43048 \ atm \approx 2.43 \ atm$.
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DifficultMCQ
At $0^{\circ} C$,a urea solution has an osmotic pressure of $400 \ mm$. On dilution by $x$ times,its osmotic pressure decreases to $100 \ mm$ at $20^{\circ} C$. The dilution factor $x$ is approximately
A
$4.3$
B
$2$
C
$5$
D
$6.8$
Solution
(A) The osmotic pressure formula is $\pi = CRT$,where $C$ is the molar concentration $(n/V)$.
For the initial state: $\pi_1 = C_1 RT_1$,where $\pi_1 = 400 \ mm$ and $T_1 = 273 \ K$.
For the final state: $\pi_2 = C_2 RT_2$,where $\pi_2 = 100 \ mm$ and $T_2 = 293 \ K$.
Taking the ratio: $\frac{\pi_1}{\pi_2} = \frac{C_1 T_1}{C_2 T_2}$.
Since $C_1/C_2 = V_2/V_1 = x$,we have $\frac{400}{100} = x \times \frac{273}{293}$.
Solving for $x$: $x = 4 \times \frac{293}{273} \approx 4 \times 1.073 = 4.292$.
Thus,the dilution factor $x$ is approximately $4.3$.
For the initial state: $\pi_1 = C_1 RT_1$,where $\pi_1 = 400 \ mm$ and $T_1 = 273 \ K$.
For the final state: $\pi_2 = C_2 RT_2$,where $\pi_2 = 100 \ mm$ and $T_2 = 293 \ K$.
Taking the ratio: $\frac{\pi_1}{\pi_2} = \frac{C_1 T_1}{C_2 T_2}$.
Since $C_1/C_2 = V_2/V_1 = x$,we have $\frac{400}{100} = x \times \frac{273}{293}$.
Solving for $x$: $x = 4 \times \frac{293}{273} \approx 4 \times 1.073 = 4.292$.
Thus,the dilution factor $x$ is approximately $4.3$.
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EasyMCQ
What is the effect of external pressure on the osmotic pressure $(OP)$ of a solution?
A
$OP$ decreases with increase of pressure
B
$OP$ decreases initially,then increases
C
$OP$ remains nearly the same with increase/decrease of external pressure
D
$OP$ increases with increase of pressure
Solution
(C) Osmotic pressure $(OP)$ is a colligative property defined by the equation $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Since $OP$ depends primarily on the concentration of the solute particles and temperature,it is an intrinsic property of the solution.
External pressure applied to the solution does not change the concentration $(C)$ or the temperature $(T)$ of the solution significantly.
Therefore,the osmotic pressure remains nearly the same regardless of changes in external pressure.
Since $OP$ depends primarily on the concentration of the solute particles and temperature,it is an intrinsic property of the solution.
External pressure applied to the solution does not change the concentration $(C)$ or the temperature $(T)$ of the solution significantly.
Therefore,the osmotic pressure remains nearly the same regardless of changes in external pressure.
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MediumMCQ
$3 \ mL$ of solution was made by dissolving $20 \ mg$ of protein at $0^{\circ} C$. The osmotic pressure of the resulting solution is $3.8 \ torr$. The molecular weight of the protein is approximately $(g/mol)$
A
$300$
B
$3 \times 10^5$
C
$3 \times 10^4$
D
$3 \times 10^3$
Solution
(C) The osmotic pressure formula is given by $\pi = \frac{W_B \times R \times T}{V \times M_B}$.
Given values: $W_B = 20 \ mg = 0.02 \ g$, $V = 3 \ mL = 0.003 \ L$, $T = 0^{\circ} C = 273 \ K$, $\pi = 3.8 \ torr = \frac{3.8}{760} \ atm = 0.005 \ atm$.
Using $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$, we rearrange for $M_B$:
$M_B = \frac{W_B \times R \times T}{V \times \pi} = \frac{0.02 \times 0.0821 \times 273}{0.003 \times 0.005}$.
$M_B = \frac{0.448266}{0.000015} \approx 29884 \ g \ mol^{-1} \approx 3 \times 10^4 \ g \ mol^{-1}$.
Given values: $W_B = 20 \ mg = 0.02 \ g$, $V = 3 \ mL = 0.003 \ L$, $T = 0^{\circ} C = 273 \ K$, $\pi = 3.8 \ torr = \frac{3.8}{760} \ atm = 0.005 \ atm$.
Using $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$, we rearrange for $M_B$:
$M_B = \frac{W_B \times R \times T}{V \times \pi} = \frac{0.02 \times 0.0821 \times 273}{0.003 \times 0.005}$.
$M_B = \frac{0.448266}{0.000015} \approx 29884 \ g \ mol^{-1} \approx 3 \times 10^4 \ g \ mol^{-1}$.
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MediumMCQ
Correct order of osmotic pressure of the following solutions is
$(i)$ $30 \ g \ L^{-1}$ of glucose
$(ii)$ $60 \ g \ L^{-1}$ of $NH_2CONH_2$
$(iii)$ $80 \ g \ L^{-1}$ of glucose
$(iv)$ $58.5 \ g \ L^{-1}$ of $NaCl$
$(i)$ $30 \ g \ L^{-1}$ of glucose
$(ii)$ $60 \ g \ L^{-1}$ of $NH_2CONH_2$
$(iii)$ $80 \ g \ L^{-1}$ of glucose
$(iv)$ $58.5 \ g \ L^{-1}$ of $NaCl$
A
$(ii)$ $< (i) < (iii) < (iv)$
B
$(i)$ $< (iii) < (ii) < (iv)$
C
$(i)$ $< (iv) < (ii) < (iii)$
D
$(iii)$ $< (i) < (iv) < (ii)$
Solution
(B) The osmotic pressure $\pi$ is given by $\pi = iCRT = i \frac{w}{MV} RT$.
Assuming $V, R, T$ are constant,$\pi \propto i \times \frac{w}{M}$.
$(i)$ $30 \ g \ L^{-1}$ glucose: $i=1, w=30, M=180 \implies \pi \propto 1 \times \frac{30}{180} \approx 0.166$.
$(ii)$ $60 \ g \ L^{-1}$ $NH_2CONH_2$: $i=1, w=60, M=60 \implies \pi \propto 1 \times \frac{60}{60} = 1.00$.
$(iii)$ $80 \ g \ L^{-1}$ glucose: $i=1, w=80, M=180 \implies \pi \propto 1 \times \frac{80}{180} \approx 0.44$.
$(iv)$ $58.5 \ g \ L^{-1}$ $NaCl$: $i=2, w=58.5, M=58.5 \implies \pi \propto 2 \times \frac{58.5}{58.5} = 2.00$.
Comparing the values: $0.166 < 0.44 < 1.00 < 2.00$.
Thus,the order is $(i) < (iii) < (ii) < (iv)$.
Assuming $V, R, T$ are constant,$\pi \propto i \times \frac{w}{M}$.
$(i)$ $30 \ g \ L^{-1}$ glucose: $i=1, w=30, M=180 \implies \pi \propto 1 \times \frac{30}{180} \approx 0.166$.
$(ii)$ $60 \ g \ L^{-1}$ $NH_2CONH_2$: $i=1, w=60, M=60 \implies \pi \propto 1 \times \frac{60}{60} = 1.00$.
$(iii)$ $80 \ g \ L^{-1}$ glucose: $i=1, w=80, M=180 \implies \pi \propto 1 \times \frac{80}{180} \approx 0.44$.
$(iv)$ $58.5 \ g \ L^{-1}$ $NaCl$: $i=2, w=58.5, M=58.5 \implies \pi \propto 2 \times \frac{58.5}{58.5} = 2.00$.
Comparing the values: $0.166 < 0.44 < 1.00 < 2.00$.
Thus,the order is $(i) < (iii) < (ii) < (iv)$.
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DifficultMCQ
$15 \%$ aqueous solution of glucose (molecular weight $= 180 \ g/mol$) is isotonic with $8 \%$ aqueous solution containing an unknown non-dissociable solute. What is the molecular weight of the unknown solute?
A
$108$
B
$96$
C
$84$
D
$9.6$
Solution
(B) For isotonic solutions,the molar concentrations are equal: $C_1 = C_2$.
Since the solutions are aqueous,we assume $100 \ mL$ of solution for both,meaning the volume $V$ is the same.
For glucose: $w_B = 15 \ g$,$m_B = 180 \ g/mol$.
For unknown solute: $w_B = 8 \ g$,$m_B = ?$.
The formula for molarity is $M = \frac{w_B \times 1000}{m_B \times V}$.
Equating the molarities: $\frac{15}{180} = \frac{8}{m_B}$.
Solving for $m_B$: $m_B = \frac{8 \times 180}{15}$.
$m_B = 8 \times 12 = 96 \ g/mol$.
Since the solutions are aqueous,we assume $100 \ mL$ of solution for both,meaning the volume $V$ is the same.
For glucose: $w_B = 15 \ g$,$m_B = 180 \ g/mol$.
For unknown solute: $w_B = 8 \ g$,$m_B = ?$.
The formula for molarity is $M = \frac{w_B \times 1000}{m_B \times V}$.
Equating the molarities: $\frac{15}{180} = \frac{8}{m_B}$.
Solving for $m_B$: $m_B = \frac{8 \times 180}{15}$.
$m_B = 8 \times 12 = 96 \ g/mol$.
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MediumMCQ
$A$ $1.17 \%$ solution of solute $A$ is isotonic with a $7.2 \%$ solution of glucose. If the molecular weight of solute $A$ is $58.5$,the value of the van't Hoff factor,'$i$',is:
A
$2$
B
$3$
C
$4$
D
$5$
Solution
(A) For isotonic solutions,the osmotic pressure $\pi$ is equal,so $\pi_1 = \pi_2$.
Since $\pi = i \cdot C \cdot R \cdot T$,and $C = \frac{w \% \times 10}{M}$,we have $i_1 \cdot \frac{w_1 \%}{M_1} = i_2 \cdot \frac{w_2 \%}{M_2}$.
For glucose,the van't Hoff factor $i_2 = 1$ and molar mass $M_2 = 180 \ g/mol$.
Given $w_1 \% = 1.17$,$M_1 = 58.5$,and $w_2 \% = 7.2$.
Substituting the values: $i_1 \times \frac{1.17}{58.5} = 1 \times \frac{7.2}{180}$.
$i_1 \times 0.02 = 0.04$.
$i_1 = \frac{0.04}{0.02} = 2$.
Since $\pi = i \cdot C \cdot R \cdot T$,and $C = \frac{w \% \times 10}{M}$,we have $i_1 \cdot \frac{w_1 \%}{M_1} = i_2 \cdot \frac{w_2 \%}{M_2}$.
For glucose,the van't Hoff factor $i_2 = 1$ and molar mass $M_2 = 180 \ g/mol$.
Given $w_1 \% = 1.17$,$M_1 = 58.5$,and $w_2 \% = 7.2$.
Substituting the values: $i_1 \times \frac{1.17}{58.5} = 1 \times \frac{7.2}{180}$.
$i_1 \times 0.02 = 0.04$.
$i_1 = \frac{0.04}{0.02} = 2$.
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MediumMCQ
At $300 \ K$,a one litre solution of sucrose (molecular weight: $342$) was prepared by dissolving $40 \ g$ of sucrose. What is the approximate osmotic pressure (in $kPa$) of the solution at the same temperature? $(R = 8.314 \times 10^6 \ cm^3 \ Pa \ K^{-1} \ mol^{-1})$
A
$292$
B
$500$
C
$292000$
D
$600$
Solution
(A) The formula for osmotic pressure is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
First,calculate the molar concentration $C$:
$C = \frac{\text{moles of solute}}{\text{volume of solution in litres}} = \frac{40 \ g / 342 \ g \ mol^{-1}}{1 \ L} = 0.11696 \ mol \ L^{-1}$.
Given $R = 8.314 \times 10^6 \ cm^3 \ Pa \ K^{-1} \ mol^{-1}$. Since $1 \ L = 1000 \ cm^3$,$R = 8.314 \times 10^3 \ L \ Pa \ K^{-1} \ mol^{-1}$.
Now,calculate $\pi$:
$\pi = 0.11696 \ mol \ L^{-1} \times 8.314 \times 10^3 \ L \ Pa \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 291.71 \times 10^3 \ Pa = 291.71 \ kPa$.
Rounding to the nearest whole number,$\pi \approx 292 \ kPa$.
First,calculate the molar concentration $C$:
$C = \frac{\text{moles of solute}}{\text{volume of solution in litres}} = \frac{40 \ g / 342 \ g \ mol^{-1}}{1 \ L} = 0.11696 \ mol \ L^{-1}$.
Given $R = 8.314 \times 10^6 \ cm^3 \ Pa \ K^{-1} \ mol^{-1}$. Since $1 \ L = 1000 \ cm^3$,$R = 8.314 \times 10^3 \ L \ Pa \ K^{-1} \ mol^{-1}$.
Now,calculate $\pi$:
$\pi = 0.11696 \ mol \ L^{-1} \times 8.314 \times 10^3 \ L \ Pa \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 291.71 \times 10^3 \ Pa = 291.71 \ kPa$.
Rounding to the nearest whole number,$\pi \approx 292 \ kPa$.
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EasyMCQ
How many grams of glucose must be added to $0.5 \ L$ of a solution so that its osmotic pressure is same as that of a solution of $9.2 \ g$ of glucose dissolved in a liter?
A
$1.15$
B
$9.22$
C
$2.31$
D
$4.6$
Solution
(D) The osmotic pressure $\pi$ is given by the formula $\pi = CRT = \frac{w}{MV}RT$.
For two solutions to have the same osmotic pressure at the same temperature,their molar concentrations must be equal: $C_1 = C_2$.
$\frac{w_1}{M_1 V_1} = \frac{w_2}{M_2 V_2}$.
Since the solute is glucose in both cases,$M_1 = M_2 = 180 \ g/mol$.
Given $V_1 = 0.5 \ L$,$w_2 = 9.2 \ g$,and $V_2 = 1.0 \ L$.
Substituting the values: $\frac{w_1}{0.5} = \frac{9.2}{1.0}$.
$w_1 = 9.2 \times 0.5 = 4.6 \ g$.
Thus,$4.6 \ g$ of glucose must be added.
For two solutions to have the same osmotic pressure at the same temperature,their molar concentrations must be equal: $C_1 = C_2$.
$\frac{w_1}{M_1 V_1} = \frac{w_2}{M_2 V_2}$.
Since the solute is glucose in both cases,$M_1 = M_2 = 180 \ g/mol$.
Given $V_1 = 0.5 \ L$,$w_2 = 9.2 \ g$,and $V_2 = 1.0 \ L$.
Substituting the values: $\frac{w_1}{0.5} = \frac{9.2}{1.0}$.
$w_1 = 9.2 \times 0.5 = 4.6 \ g$.
Thus,$4.6 \ g$ of glucose must be added.
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EasyMCQ
The molar mass of a solute $X$ in $g \ mol^{-1}$,if its $1 \%$ solution is isotonic with a $5 \%$ solution of cane sugar (molar mass $= 342 \ g \ mol^{-1}$),is
A
$68.4$
B
$34.2$
C
$136.2$
D
$171.2$
Solution
(A) For isotonic solutions,the osmotic pressure is equal,so the molar concentrations are equal: $\frac{W_1}{M_1 V_1} = \frac{W_2}{M_2 V_2}$.
Given that the solutions are $1 \%$ and $5 \%$ by mass,we can assume $1 \ g$ of solute $X$ in $100 \ mL$ solution and $5 \ g$ of cane sugar in $100 \ mL$ solution.
Here,$W_1 = 1 \ g$,$W_2 = 5 \ g$,$M_2 = 342 \ g \ mol^{-1}$,and $V_1 = V_2 = 100 \ mL$.
Substituting the values: $\frac{1}{M_1 \times 100} = \frac{5}{342 \times 100}$.
$M_1 = \frac{342}{5} = 68.4 \ g \ mol^{-1}$.
Given that the solutions are $1 \%$ and $5 \%$ by mass,we can assume $1 \ g$ of solute $X$ in $100 \ mL$ solution and $5 \ g$ of cane sugar in $100 \ mL$ solution.
Here,$W_1 = 1 \ g$,$W_2 = 5 \ g$,$M_2 = 342 \ g \ mol^{-1}$,and $V_1 = V_2 = 100 \ mL$.
Substituting the values: $\frac{1}{M_1 \times 100} = \frac{5}{342 \times 100}$.
$M_1 = \frac{342}{5} = 68.4 \ g \ mol^{-1}$.
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MediumMCQ
Which one of the following is an isotonic pair of solutions?
A
$0.15 \ M \ NaCl$ and $0.1 \ M \ Na_2SO_4$
B
$0.2 \ M$ Urea and $0.1 \ M$ Sugar
C
$0.1 \ M \ BaCl_2$ and $0.2 \ M$ Urea
D
$0.4 \ M \ MgSO_4$ and $0.1 \ M \ NH_4Cl$
Solution
(A) Two solutions are isotonic if they have the same molar concentration of particles (osmotic pressure).
For $0.15 \ M \ NaCl$: $NaCl$ dissociates into $2$ ions ($Na^+$ and $Cl^-$). The concentration of particles $= 0.15 \times 2 = 0.30 \ M$.
For $0.1 \ M \ Na_2SO_4$: $Na_2SO_4$ dissociates into $3$ ions ($2Na^+$ and $SO_4^{2-}$). The concentration of particles $= 0.1 \times 3 = 0.30 \ M$.
Since both solutions have the same concentration of particles $(0.30 \ M)$,they are isotonic.
For $0.15 \ M \ NaCl$: $NaCl$ dissociates into $2$ ions ($Na^+$ and $Cl^-$). The concentration of particles $= 0.15 \times 2 = 0.30 \ M$.
For $0.1 \ M \ Na_2SO_4$: $Na_2SO_4$ dissociates into $3$ ions ($2Na^+$ and $SO_4^{2-}$). The concentration of particles $= 0.1 \times 3 = 0.30 \ M$.
Since both solutions have the same concentration of particles $(0.30 \ M)$,they are isotonic.
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DifficultMCQ
$A$ solution containing $4 \ g$ of polymer in $4.0 \ L$ solution at $27^{\circ} C$ shows an osmotic pressure of $3.0 \times 10^{-4} \ atm$. The molar mass of the polymer in $g \ mol^{-1}$ is
A
$820000$
B
$82000$
C
$8200$
D
$820$
Solution
(B) The osmotic pressure formula is $\pi = CRT$,where $C$ is the molarity,$R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$,and $T$ is the temperature in Kelvin.
Given: $\pi = 3.0 \times 10^{-4} \ atm$,$T = 27 + 273 = 300 \ K$,$V = 4.0 \ L$,and mass of solute $w = 4 \ g$.
$C = \frac{\pi}{RT} = \frac{3.0 \times 10^{-4}}{0.0821 \times 300} = 1.218 \times 10^{-5} \ mol \ L^{-1}$.
Since $C = \frac{w}{M \times V}$,where $M$ is the molar mass:
$1.218 \times 10^{-5} = \frac{4}{M \times 4.0}$.
$M = \frac{4}{1.218 \times 10^{-5} \times 4.0} = \frac{1}{1.218 \times 10^{-5}} \approx 82101 \ g \ mol^{-1}$.
Rounding to the nearest given option,the molar mass is $82000 \ g \ mol^{-1}$.
Given: $\pi = 3.0 \times 10^{-4} \ atm$,$T = 27 + 273 = 300 \ K$,$V = 4.0 \ L$,and mass of solute $w = 4 \ g$.
$C = \frac{\pi}{RT} = \frac{3.0 \times 10^{-4}}{0.0821 \times 300} = 1.218 \times 10^{-5} \ mol \ L^{-1}$.
Since $C = \frac{w}{M \times V}$,where $M$ is the molar mass:
$1.218 \times 10^{-5} = \frac{4}{M \times 4.0}$.
$M = \frac{4}{1.218 \times 10^{-5} \times 4.0} = \frac{1}{1.218 \times 10^{-5}} \approx 82101 \ g \ mol^{-1}$.
Rounding to the nearest given option,the molar mass is $82000 \ g \ mol^{-1}$.
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MediumMCQ
Assuming the compounds to be completely dissociated in aqueous solution,identify the pair of the solutions that can be expected to be isotonic at the same temperature.
A
$0.01 \ M$ urea and $0.01 \ M \ NaCl$
B
$0.02 \ M \ NaCl$ and $0.01 \ M \ Na_2SO_4$
C
$0.03 \ M \ NaCl$ and $0.02 \ M \ MgCl_2$
D
$0.01 \ M$ sucrose and $0.02 \ M$ glucose
Solution
(C) For isotonic solutions,the effective concentration (osmolarity) must be the same: $i_1C_1 = i_2C_2$.
Assuming complete dissociation:
For $0.03 \ M \ NaCl$: $i = 2$,so $iC = 2 \times 0.03 = 0.06 \ M$.
For $0.02 \ M \ MgCl_2$: $i = 3$,so $iC = 3 \times 0.02 = 0.06 \ M$.
Since the $iC$ values are equal,the solutions are isotonic.
Assuming complete dissociation:
For $0.03 \ M \ NaCl$: $i = 2$,so $iC = 2 \times 0.03 = 0.06 \ M$.
For $0.02 \ M \ MgCl_2$: $i = 3$,so $iC = 3 \times 0.02 = 0.06 \ M$.
Since the $iC$ values are equal,the solutions are isotonic.
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MediumMCQ
At a certain temperature,the value of the slope of the plot of osmotic pressure $(\pi)$ against concentration $(C \text{ in } mol \ L^{-1})$ of a certain polymer solution is $291 \ R.$ The temperature at which osmotic pressure is measured is: $(R \text{ is the gas constant})$
A
$271^{\circ} C$
B
$18^{\circ} C$
C
$564 \ K$
D
$18 \ K$
Solution
(B) The osmotic pressure $(\pi)$ is given by the equation: $\pi = CRT.$
Comparing this with the equation of a straight line $y = mx,$ where $y = \pi,$ $x = C,$ and the slope $m = RT.$
Given that the slope is $291 \ R,$
$RT = 291 \ R$
$T = 291 \ K$
To convert the temperature to Celsius: $T(^{\circ} C) = 291 - 273 = 18^{\circ} C.$
Comparing this with the equation of a straight line $y = mx,$ where $y = \pi,$ $x = C,$ and the slope $m = RT.$
Given that the slope is $291 \ R,$
$RT = 291 \ R$
$T = 291 \ K$
To convert the temperature to Celsius: $T(^{\circ} C) = 291 - 273 = 18^{\circ} C.$
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DifficultMCQ
$A$ solution is prepared by dissolving $0.3 \ g$ of a non-volatile non-electrolyte solute '$A$' of molar mass $60 \ g \ mol^{-1}$ and $0.9 \ g$ of a non-volatile non-electrolyte solute '$B$' of molar mass $180 \ g \ mol^{-1}$ in $100 \ mL$ $H_2O$ at $27^{\circ}C$. Osmotic pressure of the solution will be
[Given: $R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}$] (in $atm$)
[Given: $R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}$] (in $atm$)
A
$1.23$
B
$2.46$
C
$0.82$
D
$1.47$
Solution
(B) Moles of solute '$A$' = $\frac{0.3 \ g}{60 \ g \ mol^{-1}} = 0.005 \ mol$.
Moles of solute '$B$' = $\frac{0.9 \ g}{180 \ g \ mol^{-1}} = 0.005 \ mol$.
Total moles of solute = $0.005 + 0.005 = 0.01 \ mol$.
Volume of solution = $100 \ mL = 0.1 \ L$.
Molarity $(C)$ = $\frac{\text{Total moles}}{\text{Volume in } L} = \frac{0.01 \ mol}{0.1 \ L} = 0.1 \ M$.
Temperature $(T)$ = $27 + 273 = 300 \ K$.
Using the formula for osmotic pressure: $\pi = CRT$.
$\pi = 0.1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 2.46 \ atm$.
Moles of solute '$B$' = $\frac{0.9 \ g}{180 \ g \ mol^{-1}} = 0.005 \ mol$.
Total moles of solute = $0.005 + 0.005 = 0.01 \ mol$.
Volume of solution = $100 \ mL = 0.1 \ L$.
Molarity $(C)$ = $\frac{\text{Total moles}}{\text{Volume in } L} = \frac{0.01 \ mol}{0.1 \ L} = 0.1 \ M$.
Temperature $(T)$ = $27 + 273 = 300 \ K$.
Using the formula for osmotic pressure: $\pi = CRT$.
$\pi = 0.1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 300 \ K$.
$\pi = 2.46 \ atm$.
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View Solution316
DifficultMCQ
The osmotic pressure of a living cell is $12 \ atm$ at $300 \ K$. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is $...... \ g \ L^{-1}$. (Nearest integer)
Given : $R = 0.08 \ L \ atm \ K^{-1} \ mol^{-1}$
Assume complete dissociation of $NaCl$
(Given : Molar mass of $Na$ and $Cl$ are $23$ and $35.5 \ g \ mol^{-1}$ respectively.)
Given : $R = 0.08 \ L \ atm \ K^{-1} \ mol^{-1}$
Assume complete dissociation of $NaCl$
(Given : Molar mass of $Na$ and $Cl$ are $23$ and $35.5 \ g \ mol^{-1}$ respectively.)
A
$15$
B
$30$
C
$7.5$
D
$58.5$
Solution
(A) For an isotonic solution,the osmotic pressure $(\pi)$ of the $NaCl$ solution must be equal to the osmotic pressure of the living cell.
$\pi = iCRT$
Given $\pi = 12 \ atm$,$T = 300 \ K$,$R = 0.08 \ L \ atm \ K^{-1} \ mol^{-1}$,and $i = 2$ (for $NaCl \rightarrow Na^{+} + Cl^{-}$).
$12 = 2 \times C \times 0.08 \times 300$
$12 = 48 \times C$
$C = \frac{12}{48} = 0.25 \ mol \ L^{-1}$
The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g \ mol^{-1}$.
Strength of $NaCl$ solution = $C \times \text{Molar mass}$
$= 0.25 \times 58.5 = 14.625 \ g \ L^{-1}$.
Rounding to the nearest integer,we get $15 \ g \ L^{-1}$.
$\pi = iCRT$
Given $\pi = 12 \ atm$,$T = 300 \ K$,$R = 0.08 \ L \ atm \ K^{-1} \ mol^{-1}$,and $i = 2$ (for $NaCl \rightarrow Na^{+} + Cl^{-}$).
$12 = 2 \times C \times 0.08 \times 300$
$12 = 48 \times C$
$C = \frac{12}{48} = 0.25 \ mol \ L^{-1}$
The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g \ mol^{-1}$.
Strength of $NaCl$ solution = $C \times \text{Molar mass}$
$= 0.25 \times 58.5 = 14.625 \ g \ L^{-1}$.
Rounding to the nearest integer,we get $15 \ g \ L^{-1}$.
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View Solution317
MediumMCQ
. . . . . . solution is hypertonic with reference to fluid inside the blood cell.
A
$0.8\% \text{ W/V NaCl}$
B
$0.6\% \text{ W/V NaCl}$
C
$0.9\% \text{ W/V NaCl}$
D
$1.2\% \text{ W/V NaCl}$
Solution
(D) The isotonic concentration for human red blood cells is $0.9\% \text{ W/V NaCl}$.
Any solution with a concentration higher than $0.9\% \text{ W/V NaCl}$ is considered hypertonic with respect to the fluid inside the blood cell.
Among the given options,$1.2\% \text{ W/V NaCl}$ has a higher concentration than $0.9\% \text{ W/V NaCl}$.
Therefore,$1.2\% \text{ W/V NaCl}$ is hypertonic. Option $(D)$ is correct.
Any solution with a concentration higher than $0.9\% \text{ W/V NaCl}$ is considered hypertonic with respect to the fluid inside the blood cell.
Among the given options,$1.2\% \text{ W/V NaCl}$ has a higher concentration than $0.9\% \text{ W/V NaCl}$.
Therefore,$1.2\% \text{ W/V NaCl}$ is hypertonic. Option $(D)$ is correct.
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View Solution318
DifficultMCQ
Given below are two statements:
Statement $I$: $H_2O$ molecules move from the chamber $1$ to chamber $2$.
Statement $II$: The osmotic pressure of a solution prepared by dissolving $50 \text{ mg}$ of potassium sulphate (molar mass = $174 \text{ g/mol}$) in $2 \text{ L}$ of water (at $27^{\circ}C$) is $0.0107 \text{ bar}$. (Given: $R = 0.083 \text{ dm}^3 \text{ bar K}^{-1} \text{ mol}^{-1}$ and assume complete dissociation of electrolyte).
In the light of the above statements,choose the correct answer from the options given below:
Statement $I$: $H_2O$ molecules move from the chamber $1$ to chamber $2$.
Statement $II$: The osmotic pressure of a solution prepared by dissolving $50 \text{ mg}$ of potassium sulphate (molar mass = $174 \text{ g/mol}$) in $2 \text{ L}$ of water (at $27^{\circ}C$) is $0.0107 \text{ bar}$. (Given: $R = 0.083 \text{ dm}^3 \text{ bar K}^{-1} \text{ mol}^{-1}$ and assume complete dissociation of electrolyte).
In the light of the above statements,choose the correct answer from the options given below:
A
Both Statement $I$ and Statement $II$ are true
B
Both Statement $I$ and Statement $II$ are false
C
Statement $I$ is true but Statement $II$ is false
D
Statement $I$ is false but Statement $II$ is true
Solution
(D) For Statement $I$:
Chamber $1$ contains $18 \text{ g}$ of glucose in $100 \text{ mL}$ of solution. Molarity $(M_1)$ = $\frac{18/180}{0.1} = 1.0 \text{ M}$.
Chamber $2$ contains $30 \text{ g}$ of glucose in $250 \text{ mL}$ of solution. Molarity $(M_2)$ = $\frac{30/180}{0.25} = 0.667 \text{ M}$.
Since solvent moves from a region of lower solute concentration to higher solute concentration through a semi-permeable membrane,water moves from chamber $2$ to chamber $1$. Thus,Statement $I$ is false.
For Statement $II$:
Potassium sulphate $(K_2SO_4)$ dissociates as $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$,so the van't Hoff factor $(i)$ = $3$.
Given: mass = $50 \text{ mg} = 0.05 \text{ g}$,molar mass = $174 \text{ g/mol}$,volume $(V)$ = $2 \text{ L}$,$T = 27 + 273 = 300 \text{ K}$,$R = 0.083 \text{ dm}^3 \text{ bar K}^{-1} \text{ mol}^{-1}$.
Osmotic pressure $(\pi)$ = $i \times C \times R \times T = i \times (n/V) \times R \times T = 3 \times (0.05 / 174) / 2 \times 0.083 \times 300 = 0.0107 \text{ bar}$.
Thus,Statement $II$ is true.
Chamber $1$ contains $18 \text{ g}$ of glucose in $100 \text{ mL}$ of solution. Molarity $(M_1)$ = $\frac{18/180}{0.1} = 1.0 \text{ M}$.
Chamber $2$ contains $30 \text{ g}$ of glucose in $250 \text{ mL}$ of solution. Molarity $(M_2)$ = $\frac{30/180}{0.25} = 0.667 \text{ M}$.
Since solvent moves from a region of lower solute concentration to higher solute concentration through a semi-permeable membrane,water moves from chamber $2$ to chamber $1$. Thus,Statement $I$ is false.
For Statement $II$:
Potassium sulphate $(K_2SO_4)$ dissociates as $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$,so the van't Hoff factor $(i)$ = $3$.
Given: mass = $50 \text{ mg} = 0.05 \text{ g}$,molar mass = $174 \text{ g/mol}$,volume $(V)$ = $2 \text{ L}$,$T = 27 + 273 = 300 \text{ K}$,$R = 0.083 \text{ dm}^3 \text{ bar K}^{-1} \text{ mol}^{-1}$.
Osmotic pressure $(\pi)$ = $i \times C \times R \times T = i \times (n/V) \times R \times T = 3 \times (0.05 / 174) / 2 \times 0.083 \times 300 = 0.0107 \text{ bar}$.
Thus,Statement $II$ is true.
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View Solution319
MediumMCQ
At $27^{\circ}C$,$0.1 \ M$,$1 \ L$ $K_4[Fe(CN)_6]$ aqueous solution and $0.1 \ M$,$1 \ L$ $FeCl_3$ aqueous solution are placed in a container separated by a semi-permeable membrane $AB$. Assume complete dissociation of both the solutes. Which of the following statements is correct?
A
Blue color is formed on both sides.
B
Ionic solutes in aqueous solution can pass through semi-permeable membrane.
C
Solution on side 'y' is hypotonic.
D
To cause the reverse flow of solvent during osmosis,external pressure (any value) should be applied to side 'x'.
Solution
(C) Side $x$: $K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$. The van't Hoff factor $i = 5$. The effective concentration is $0.1 \ M \times 5 = 0.5 \ M$.
Side $y$: $FeCl_3 \rightarrow Fe^{3+} + 3Cl^-$. The van't Hoff factor $i = 4$. The effective concentration is $0.1 \ M \times 4 = 0.4 \ M$.
Osmotic pressure is given by $\pi = iCRT$. Since the temperature and concentration of the solute are the same,the osmotic pressure depends on the van't Hoff factor $i$.
Since $i_x > i_y$,the osmotic pressure of side $x$ is greater than that of side $y$ $(\pi_x > \pi_y)$.
Therefore,the solution on side $y$ has a lower osmotic pressure and is considered hypotonic relative to side $x$.
Side $y$: $FeCl_3 \rightarrow Fe^{3+} + 3Cl^-$. The van't Hoff factor $i = 4$. The effective concentration is $0.1 \ M \times 4 = 0.4 \ M$.
Osmotic pressure is given by $\pi = iCRT$. Since the temperature and concentration of the solute are the same,the osmotic pressure depends on the van't Hoff factor $i$.
Since $i_x > i_y$,the osmotic pressure of side $x$ is greater than that of side $y$ $(\pi_x > \pi_y)$.
Therefore,the solution on side $y$ has a lower osmotic pressure and is considered hypotonic relative to side $x$.
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View Solution320
DifficultMCQ
Solution $A$ is prepared by dissolving $1 \text{ g}$ of a protein (molar mass = $50000 \text{ g mol}^{-1}$) in $0.5 \text{ L}$ of water at $300 \text{ K}$. Its osmotic pressure is $x \text{ bar}$. Solution $B$ is made by dissolving $2 \text{ g}$ of the same protein in $1 \text{ L}$ of water at $300 \text{ K}$. The osmotic pressure of solution $B$ is $y \text{ bar}$. The entire solution of $A$ is mixed with the entire solution of $B$ at the same temperature. The osmotic pressure of the resultant solution is $z \text{ bar}$. $x, y$ and $z$ respectively are: $(R = 0.083 \text{ L bar mol}^{-1} \text{ K}^{-1})$
A
$9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}$
B
$9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 19.92 \times 10^{-4}$
C
$4.98 \times 10^{-4}; 4.98 \times 10^{-4}; 9.96 \times 10^{-4}$
D
$4.98 \times 10^{-4}; 4.98 \times 10^{-4}; 4.98 \times 10^{-4}$
Solution
(A) The formula for osmotic pressure is $\Pi = (n/V)RT$.
For solution $A$: $n_A = 1 \text{ g} / 50000 \text{ g mol}^{-1} = 2 \times 10^{-5} \text{ mol}$. $V_A = 0.5 \text{ L}$.
$x = (2 \times 10^{-5} \text{ mol} / 0.5 \text{ L}) \times 0.083 \text{ L bar mol}^{-1} \text{ K}^{-1} \times 300 \text{ K} = 4 \times 10^{-5} \times 24.9 = 9.96 \times 10^{-4} \text{ bar}$.
For solution $B$: $n_B = 2 \text{ g} / 50000 \text{ g mol}^{-1} = 4 \times 10^{-5} \text{ mol}$. $V_B = 1 \text{ L}$.
$y = (4 \times 10^{-5} \text{ mol} / 1 \text{ L}) \times 0.083 \times 300 = 9.96 \times 10^{-4} \text{ bar}$.
For the mixed solution: $n_{tot} = n_A + n_B = (2 + 4) \times 10^{-5} = 6 \times 10^{-5} \text{ mol}$.
$V_{tot} = V_A + V_B = 0.5 \text{ L} + 1 \text{ L} = 1.5 \text{ L}$.
$z = (6 \times 10^{-5} \text{ mol} / 1.5 \text{ L}) \times 24.9 = 4 \times 10^{-5} \times 24.9 = 9.96 \times 10^{-4} \text{ bar}$.
Thus,$x = y = z = 9.96 \times 10^{-4}$.
For solution $A$: $n_A = 1 \text{ g} / 50000 \text{ g mol}^{-1} = 2 \times 10^{-5} \text{ mol}$. $V_A = 0.5 \text{ L}$.
$x = (2 \times 10^{-5} \text{ mol} / 0.5 \text{ L}) \times 0.083 \text{ L bar mol}^{-1} \text{ K}^{-1} \times 300 \text{ K} = 4 \times 10^{-5} \times 24.9 = 9.96 \times 10^{-4} \text{ bar}$.
For solution $B$: $n_B = 2 \text{ g} / 50000 \text{ g mol}^{-1} = 4 \times 10^{-5} \text{ mol}$. $V_B = 1 \text{ L}$.
$y = (4 \times 10^{-5} \text{ mol} / 1 \text{ L}) \times 0.083 \times 300 = 9.96 \times 10^{-4} \text{ bar}$.
For the mixed solution: $n_{tot} = n_A + n_B = (2 + 4) \times 10^{-5} = 6 \times 10^{-5} \text{ mol}$.
$V_{tot} = V_A + V_B = 0.5 \text{ L} + 1 \text{ L} = 1.5 \text{ L}$.
$z = (6 \times 10^{-5} \text{ mol} / 1.5 \text{ L}) \times 24.9 = 4 \times 10^{-5} \times 24.9 = 9.96 \times 10^{-4} \text{ bar}$.
Thus,$x = y = z = 9.96 \times 10^{-4}$.
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View Solution321
DifficultMCQ
$20 \text{ g}$ hemoglobin in a $1 \text{ L}$ aqueous solution $(A)$ at $300 \text{ K}$ is separated from pure water by a semi-permeable membrane. At equilibrium,the height of the solution in a tube dipped in solution $(A)$ is found to be $80.0 \text{ mm}$ higher than the tube dipped in water. The molar mass of hemoglobin is . . . . . . $\text{kg mol}^{-1}$. (Nearest integer) (Given: $g = 10 \text{ m s}^{-2}$,$R = 8.3 \text{ kPa dm}^3 \text{ K}^{-1} \text{mol}^{-1}$,density of solution = $1000 \text{ kg m}^{-3}$)
A
$60$
B
$62$
C
$65$
D
$68$
Solution
(B) $1$. The osmotic pressure $\pi$ is given by the hydrostatic pressure formula: $\pi = h \rho g$.
$2$. Convert height to meters: $h = 80.0 \text{ mm} = 0.08 \text{ m}$.
$3$. Calculate $\pi$: $\pi = 0.08 \text{ m} \times 1000 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} = 800 \text{ Pa} = 0.8 \text{ kPa}$.
$4$. Use the osmotic pressure formula: $\pi = CRT = (n/V)RT$,where $n = \text{mass}/M$.
$5$. Substitute the values: $0.8 = (20 / M) / 1 \times 8.3 \times 300$.
$6$. Solve for $M$: $M = (20 \times 8.3 \times 300) / 0.8 = 62250 \text{ g mol}^{-1} = 62.25 \text{ kg mol}^{-1}$.
$7$. The nearest integer is $62$.
$2$. Convert height to meters: $h = 80.0 \text{ mm} = 0.08 \text{ m}$.
$3$. Calculate $\pi$: $\pi = 0.08 \text{ m} \times 1000 \text{ kg m}^{-3} \times 10 \text{ m s}^{-2} = 800 \text{ Pa} = 0.8 \text{ kPa}$.
$4$. Use the osmotic pressure formula: $\pi = CRT = (n/V)RT$,where $n = \text{mass}/M$.
$5$. Substitute the values: $0.8 = (20 / M) / 1 \times 8.3 \times 300$.
$6$. Solve for $M$: $M = (20 \times 8.3 \times 300) / 0.8 = 62250 \text{ g mol}^{-1} = 62.25 \text{ kg mol}^{-1}$.
$7$. The nearest integer is $62$.
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