Method of expressing concentration of solution Questions in English

(D) $5 \ M$ molar solution means there are $5 \ mol$ of $KCl$ salt in $1 \ L$ of solution.
Molecular weight of $KCl = 74.55 \ g \ mol^{-1}$.
Hence,the mass of solute in $1 \ L$ of solution is $5 \ mol \times 74.55 \ g \ mol^{-1} = 372.75 \ g$.
$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$.
Mass of $1 \ L$ $(1000 \ mL)$ of solution $= 1.26 \ g \ mL^{-1} \times 1000 \ mL = 1260 \ g$.
Mass of solvent (water) $=$ Mass of solution $-$ Mass of solute $= 1260 \ g - 372.75 \ g = 887.25 \ g = 0.88725 \ kg$.
$\text{Molality} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in } kg} = \frac{5 \ mol}{0.88725 \ kg} \approx 5.636 \ m$.

(C) $1$ molal solution means $1$ mole of solute is dissolved in $1000 \ g$ $(1 \ kg)$ of ethanol.
Molecular weight of ethanol $(C_2H_5OH)$ = $(2 \times 12.01) + (6 \times 1.008) + 16.00 = 46.07 \ g/mol$.
Number of moles of ethanol = $\frac{1000 \ g}{46.07 \ g/mol} \approx 21.70 \ mol$.
Mole fraction of solute $(x_{solute})$ = $\frac{n_{solute}}{n_{solute} + n_{solvent}}$.
$x_{solute} = \frac{1}{1 + 21.70} = \frac{1}{22.70} \approx 0.044$.
Therefore,the mole fraction of the solute is $0.044$.

(D) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Step $1$: Calculate the number of moles of $NaOH$.
Moles of $NaOH = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{10 \ g}{40 \ g \ mol^{-1}} = 0.25 \ mol$.
Step $2$: Convert the volume of the solution to liters.
Volume $= 500 \ mL = 0.5 \ L$.
Step $3$: Calculate Molarity.
Molarity $= \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{0.25 \ mol}{0.5 \ L} = 0.5 \ M$.

(C) The relationship between molarity $(M)$,molality $(m)$,and density $(d)$ is given by the formula:
$m = \frac{1000 \times M}{1000 \times d - M \times M_B}$
Rearranging for density $(d)$:
$d = \frac{M}{m} + \frac{M \times M_B}{1000} = \frac{1.44}{1.5} + \frac{1.44 \times 166}{1000}$
$d = 0.96 + 0.23904 = 1.19904 \ g \ mL^{-1} \approx 1.20 \ g \ mL^{-1}$
Here,$M = 1.44 \ mol \ L^{-1}$,$m = 1.5 \ mol \ kg^{-1}$,and $M_B (KI) = 166 \ g \ mol^{-1}$.

(B) $PPM$ (Parts Per Million) is defined as the mass of the solute in $mg$ per $kg$ of the solution.
Given,mass of toothpaste $= 500 \ g = 0.5 \ kg$.
Mass of fluorine $= 0.2 \ g = 200 \ mg$.
Concentration in $ppm = \frac{\text{mass of solute in } mg}{\text{mass of solution in } kg} = \frac{200 \ mg}{0.5 \ kg} = 400 \ ppm$.
$400 \ ppm$ can be written as $4 \times 10^2 \ ppm$.
Therefore,option $(B)$ is correct.

(B) Given: Molality $(m) = 5 \ mol / kg$,Density $(d) = 1.3 \ g / mL$,Molar mass of urea $(M_w) = 60.06 \ g / mol$.
$5$ molal solution means $5$ moles of urea are present in $1000 \ g$ of solvent (water).
Mass of urea = $5 \ mol \times 60.06 \ g / mol = 300.3 \ g$.
Total mass of solution = Mass of solute + Mass of solvent = $300.3 \ g + 1000 \ g = 1300.3 \ g$.
Volume of solution = $\frac{\text{Mass}}{\text{Density}} = \frac{1300.3 \ g}{1.3 \ g / mL} \approx 1000.23 \ mL = 1.00023 \ L$.
Molarity $(M) = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{5 \ mol}{1.00023 \ L} \approx 4.9988 \ M$.
Since $4.9988 \ M$ is approximately $5 \ M$,the correct option is $(B)$.

(B) Mass $\%$ of solute = $\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
Mass of solute = $1 \ g$
Mass of solvent = $18 \ g$
Mass of solution = $1 \ g + 18 \ g = 19 \ g$
$\therefore$ Mass $\%$ of solute = $\frac{1}{19} \times 100 \approx 5.26 \%$
Rounding to the nearest whole number,we get $5 \%$.
Hence,the correct option is $(B)$.

(C) Given,mass of $HCl = 18.25 \ g$.
Molar mass of $HCl = 36.5 \ g/mol$.
Mass of water $= 500 \ g = 0.5 \ kg$.
Number of moles of $HCl = \frac{\text{Mass of } HCl}{\text{Molar mass of } HCl} = \frac{18.25}{36.5} = 0.5 \ mol$.
Molality $= \frac{\text{Number of moles of solute}}{\text{Mass of solvent in } kg}$.
Molality $= \frac{0.5 \ mol}{0.5 \ kg} = 1 \ m$.
Hence,the molality of the solution is $1 \ m$,thus the correct option is $C$.

(C) Mole fraction is a unit of concentration,defined as the ratio of the number of moles of a specific component to the total number of moles of all components in the solution.
Since it is a ratio of two quantities with the same unit (moles),the units cancel out,making mole fraction a unitless quantity.
The sum of the mole fractions of all components in a solution is always equal to $1$.
Therefore,the correct option is $(C)$.

(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given: Molality $(m)$ = $0.25 \ mol/kg$,Mass of solution = $2.5 \ kg$.
Let the mass of urea be $w \ g$. Molar mass of urea $(NH_2CONH_2)$ = $60 \ g/mol$.
Moles of urea $(n)$ = $\frac{w}{60}$.
Mass of solvent = Mass of solution - Mass of solute = $(2500 - w) \ g = \frac{2500 - w}{1000} \ kg$.
Using the formula: $m = \frac{n}{Mass \ of \ solvent \ (kg)}$.
$0.25 = \frac{w/60}{(2500 - w)/1000}$.
$0.25 = \frac{w}{60} \times \frac{1000}{2500 - w}$.
$0.25 = \frac{1000w}{60(2500 - w)}$.
$15(2500 - w) = 1000w$.
$37500 - 15w = 1000w$.
$1015w = 37500$.
$w \approx 36.94 \ g \approx 37 \ g$.

(C) The amount of solute in grams is calculated using the formula: $\text{Mass} = \text{Molarity} (M) \times \text{Volume} (V \text{ in } L) \times \text{Molar Mass} (MW)$.
For $A$: $\text{Mass} = 0.25 \times 1.0 \times 106 = 26.5 \ g$.
For $B$: $\text{Mass} = 0.2 \times 0.25 \times 142 = 7.1 \ g$.
For $C$: $\text{Mass} = 1.0 \times 0.5 \times 158 = 79.0 \ g$.
For $D$: $\text{Mass} = 0.5 \times 0.75 \times 60 = 22.5 \ g$.
Comparing the values,$79.0 \ g$ is the highest amount. Therefore,option $C$ is correct.

(A) The molarity $(M)$ is defined as the number of moles of solute per liter of solution.
First,calculate the moles of $NaOH$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \ g}{40 \ g/mol} = 0.05 \ mol$.
At $25^{\circ} C$,the volume is $200 \ mL = 0.2 \ L$.
Thus,the molarity at $25^{\circ} C$ is $M = \frac{0.05 \ mol}{0.2 \ L} = 0.25 \ M$.
When the temperature increases to $90^{\circ} C$,the volume of the solution increases due to thermal expansion.
Since the number of moles of $NaOH$ remains constant while the volume increases,the molarity $(M)$ will decrease.
Therefore,the molarity at $90^{\circ} C$ will be less than $0.25 \ M$.

(C) Density of solution $= 1.11 \ g \ mL^{-1}$.
Molarity of solution $= 2 \ M$, which means $2 \ \text{moles}$ of solute are present in $1000 \ mL$ of solution.
Mass of solution $= \text{density} \times \text{volume} = 1.11 \ g/mL \times 1000 \ mL = 1110 \ g$.
Molar mass of ethylene glycol $(C_2H_6O_2) = 62 \ g/mol$.
Mass of solute $= \text{moles} \times \text{molar mass} = 2 \ mol \times 62 \ g/mol = 124 \ g$.
Mass of solvent $= \text{Mass of solution} - \text{Mass of solute} = 1110 \ g - 124 \ g = 986 \ g$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{2 \ mol}{0.986 \ kg} \approx 2.02 \ m$.

(B) Given: $98\% (w/w)$ $H_2SO_4$ means $98 \ g$ of $H_2SO_4$ in $100 \ g$ of solution.
Mass of solute $(H_2SO_4) = 98 \ g$.
Molar mass of $H_2SO_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \ g/mol$.
Moles of $H_2SO_4 = \frac{98 \ g}{98 \ g/mol} = 1 \ mol$.
Mass of solvent (water) = Total mass - Mass of solute = $100 \ g - 98 \ g = 2 \ g = 0.002 \ kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1 \ mol}{0.002 \ kg} = 500 \ m$.

(B) The number of moles of $NaCl$ is $n_{NaCl} = 0.1 \ mol$.
The mass of water is $100 \ g$. The molar mass of water $(H_2O)$ is $18 \ g/mol$.
The number of moles of water is $n_{H_2O} = \frac{100 \ g}{18 \ g/mol} \approx 5.556 \ mol$.
The mole fraction of $NaCl$ $(X_{NaCl})$ is given by the formula:
$X_{NaCl} = \frac{n_{NaCl}}{n_{NaCl} + n_{H_2O}} = \frac{0.1}{0.1 + 5.556} = \frac{0.1}{5.656} \approx 0.0177$.

(D) Let $w_A$ be the mass of water $(H_2O)$ and $w_B$ be the mass of ethanol $(C_2H_5OH)$.
Given: $w_B = 69 \ g$,molar mass of ethanol $(m_B)$ = $46 \ g/mol$,molar mass of water $(m_A)$ = $18 \ g/mol$.
Mole fraction of ethanol $(X_B)$ = $0.6$.
Therefore,mole fraction of water $(X_A)$ = $1 - 0.6 = 0.4$.
The formula for mole fraction is $X_B = \frac{n_B}{n_A + n_B}$,where $n$ is the number of moles.
$n_B = \frac{69}{46} = 1.5 \ mol$.
$n_A = \frac{x}{18} \ mol$.
$0.6 = \frac{1.5}{\frac{x}{18} + 1.5}$.
$0.6 \times (\frac{x}{18} + 1.5) = 1.5$.
$\frac{0.6x}{18} + 0.9 = 1.5$.
$\frac{x}{30} = 0.6$.
$x = 0.6 \times 30 = 18 \ g$.

(B) Given,
Moles of solute $(n) = 1 \ mol$
Mass of solvent $(w_A) = 50 \ g$
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Formula: $m = \frac{n \times 1000}{w_A \text{ (in } g)}$
Substituting the values: $m = \frac{1 \times 1000}{50} = 20 \ mol \ kg^{-1}$
Hence,option $(B)$ is the correct answer.

(B) Given:
Mass of $NaCl = 13.50 \ g$
Volume of solution $= 452 \ mL = 0.452 \ L$
Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g \ mol^{-1}$
Molarity $(M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in } L}$
Number of moles of $NaCl = \frac{13.50 \ g}{58.5 \ g \ mol^{-1}} \approx 0.2308 \ mol$
$M = \frac{0.2308 \ mol}{0.452 \ L} \approx 0.51 \ mol \ L^{-1}$

(A) The concentration in $ppm$ is calculated as:
$\text{Concentration (ppm)} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6$
Given:
$\text{Mass of solute (dissolved oxygen)} = 6 \times 10^{-3} \ g$
$\text{Mass of solution (seawater)} = 1030 \ g$
$\text{Concentration} = \frac{6 \times 10^{-3}}{1030} \times 10^6$
$\text{Concentration} = \frac{6000}{1030} \approx 5.825 \ ppm$
Rounding to one decimal place, we get $5.8 \ ppm$.

(D) Given: $93 \%$ $w/V$ $H_2SO_4$ solution means $93 \ g$ of $H_2SO_4$ is present in $100 \ mL$ of solution.
For $1 \ L$ $(1000 \ mL)$ of solution,the mass of $H_2SO_4$ solute is $930 \ g$.
The density of the solution is $d = 1.84 \ g/mL$.
Total mass of $1 \ L$ solution = $Volume \times Density = 1000 \ mL \times 1.84 \ g/mL = 1840 \ g$.
Mass of solvent (water) = $\text{Total mass of solution} - \text{Mass of solute} = 1840 \ g - 930 \ g = 910 \ g = 0.91 \ kg$.
Moles of $H_2SO_4$ = $\frac{930 \ g}{98 \ g/mol} \approx 9.49 \ mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{9.49 \ mol}{0.91 \ kg} \approx 10.42 \ mol/kg$.

(A) Given: Weight of $H_2SO_4 = 98 \ g$,Molar mass of $H_2SO_4 = 98 \ g \ mol^{-1}$,Weight of solvent (water) $= 2 \ g$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Moles of $H_2SO_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{98 \ g}{98 \ g \ mol^{-1}} = 1 \ mol$.
Weight of solvent in $kg = \frac{2 \ g}{1000} = 0.002 \ kg$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{1 \ mol}{0.002 \ kg} = 500 \ m$.

(C) For $HCl$,the normality $(N)$ is equal to the molarity $(M)$ because the n-factor is $1$. Thus,$2 \ N = 2 \ M$ and $5 \ N = 5 \ M$.
Let $V_1$ be the volume of $2 \ M \ HCl$ used and $V_2$ be the volume of $5 \ M \ HCl$ used.
We are given $V_1 = 500 \ mL$ and we have a maximum of $500 \ mL$ of $5 \ M \ HCl$ available $(V_2 \leq 500 \ mL)$.
The mixture equation is $M_1 V_1 + M_2 V_2 = M_{final} (V_1 + V_2)$.
Substituting the values: $2(500) + 5(V_2) = 3(500 + V_2)$.
$1000 + 5V_2 = 1500 + 3V_2$.
$2V_2 = 500$,which gives $V_2 = 250 \ mL$.
Since $250 \ mL < 500 \ mL$,this is feasible.
The total volume of $3 \ M \ HCl$ prepared is $V_1 + V_2 = 500 \ mL + 250 \ mL = 750 \ mL$.

(B) Molarity $(M)$ = $0.5 \text{ M}$,Volume of solution = $100 \text{ cc} = 0.1 \text{ L}$.
Moles of ethyl alcohol = $M \times V = 0.5 \times 0.1 = 0.05 \text{ mol}$.
Molar mass of ethyl alcohol $(C_2H_5OH)$ = $46 \text{ g/mol}$.
Mass of ethyl alcohol = $0.05 \times 46 = 2.3 \text{ g}$.
Density = $1.15 \text{ g/cc}$.
Volume = $\frac{\text{Mass}}{\text{Density}} = \frac{2.3}{1.15} = 2 \text{ cc}$.

(B) Step $1$: Calculate the number of moles of the solute.
Moles $= \frac{\text{Mass}}{\text{Molar Mass}} = \frac{63 \ g}{126 \ g/mol} = 0.5 \ mol$.
Step $2$: Calculate the total mass of the solution.
Mass of solution $= \text{Mass of solute} + \text{Mass of solvent} = 63 \ g + 500 \ g = 563 \ g$.
Step $3$: Calculate the volume of the solution using density.
Volume $= \frac{\text{Mass}}{\text{Density}} = \frac{563 \ g}{1.126 \ g/mL} = 500 \ mL = 0.5 \ L$.
Step $4$: Calculate the molarity.
Molarity $(M) = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{0.5 \ mol}{0.5 \ L} = 1.0 \ M$.

(A) Given:
Mole fraction of solute $(x_2)$ = $0.1$
Mole fraction of solvent $(x_1)$ = $1 - 0.1 = 0.9$
Density $(d)$ = $2.0 \ g \ cm^{-3}$
Molarity $(M)$ = Molality $(m)$
We know that:
$m = \frac{x_2 \times 1000}{x_1 \times M_1}$
$M = \frac{1000 \times d \times x_2}{x_1 M_1 + x_2 M_2}$
Since $M = m$:
$\frac{x_2 \times 1000}{x_1 \times M_1} = \frac{1000 \times d \times x_2}{x_1 M_1 + x_2 M_2}$
$\frac{1}{x_1 M_1} = \frac{d}{x_1 M_1 + x_2 M_2}$
$x_1 M_1 + x_2 M_2 = d \times x_1 M_1$
$0.9 M_1 + 0.1 M_2 = 2.0 \times 0.9 M_1$
$0.9 M_1 + 0.1 M_2 = 1.8 M_1$
$0.1 M_2 = 0.9 M_1$
$\frac{M_2}{M_1} = \frac{0.9}{0.1} = 9$
Therefore,the ratio of molecular weights of the solute and the solvent is $9$.

(B) Given: Mole fraction of ethanol $(x_{EtOH})$ = $0.08$.
Since the sum of mole fractions is $1$,the mole fraction of water $(x_{H_2O})$ = $1 - 0.08 = 0.92$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
The formula relating molality to mole fraction is:
$m = \frac{x_{solute}}{x_{solvent}} \times \frac{1000}{MW_{solvent}}$
Substituting the values:
$m = \frac{0.08}{0.92} \times \frac{1000}{18.02} \approx 4.83 \ mol \ kg^{-1}$.

(C) Total mass of $H_2SO_4 = (100 \times 0.98) + (100 \times 0.49) = 98 + 49 = 147 \ g$.
Total mass of solution $= 100 + 100 = 200 \ g$.
Total mass of $H_2O = 200 - 147 = 53 \ g$.
Moles of $H_2SO_4 = \frac{147}{98} = 1.5 \ mol$.
Moles of $H_2O = \frac{53}{18} \approx 2.944 \ mol$.
Mole fraction of $H_2SO_4 = \frac{n_{H_2SO_4}}{n_{H_2SO_4} + n_{H_2O}} = \frac{1.5}{1.5 + 2.944} = \frac{1.5}{4.444} \approx 0.337$.

(B) Let the number of moles of $NaOH = 0.2$ and moles of $H_2O = 0.8$ (since the sum of mole fractions is $1$).
Mass of $NaOH = \text{moles} \times \text{molar mass} = 0.2 \times 40 = 8 \text{ g}$.
Mass of $H_2O = \text{moles} \times \text{molar mass} = 0.8 \times 18 = 14.4 \text{ g}$.
Mass percentage of $NaOH = \frac{\text{Mass of } NaOH}{\text{Mass of } NaOH + \text{Mass of } H_2O} \times 100$.
Mass percentage $= \frac{8}{8 + 14.4} \times 100 = \frac{8}{22.4} \times 100 \approx 35.71\%$.
Thus, option $(B)$ is correct.

(A) The molar mass of $\text{NaOH}$ is calculated as $23 + 16 + 1 = 40 \text{ g/mol}$.
Number of moles of $\text{NaOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{15 \text{ g}}{40 \text{ g/mol}} = 0.375 \text{ mol}$.
The volume of the solution is $900 \text{ ml} = 0.9 \text{ L}$.
Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
$M = \frac{0.375 \text{ mol}}{0.9 \text{ L}} = 0.4166... \text{ M} \approx 0.42 \text{ M}$.

(D) $1$. Molar mass of urea $(NH_2CONH_2) = 2(14) + 4(1) + 12 + 16 = 60 \text{ g mol}^{-1}$.
$2$. For $100 \text{ g}$ of solution,mass of urea $= 10 \text{ g}$,mass of water $= 90 \text{ g}$.
$3$. Moles of urea $= 10 / 60 = 1/6 \approx 0.1667 \text{ mol}$.
$4$. Moles of water $= 90 / 18 = 5 \text{ mol}$.
$5$. Mole fraction of water $(x_{H_2O}) = \frac{n_{H_2O}}{n_{H_2O} + n_{urea}} = \frac{5}{5 + 0.1667} = \frac{5}{5.1667} \approx 0.967$.