Method of expressing concentration of solution Questions in English

(B) Step $1$: Calculate the molarity of the initial solution.
$M_1 = \frac{W \times 1000}{MW \times V(mL)} = \frac{2.65 \times 1000}{106 \times 250} = 0.1 \, M$
Step $2$: Use the dilution formula $M_1V_1 = M_2V_2$ to find the final concentration.
$M_1 = 0.1 \, M, V_1 = 10 \, mL$
$V_2 = 1 \, L = 1000 \, mL$
$0.1 \times 10 = M_2 \times 1000$
$M_2 = \frac{1}{1000} = 0.001 \, M$

(C) The correct option is $C$.
By definition,molarity $(M)$ is defined as the number of moles of solute dissolved in one litre of the solution.
Therefore,a molar solution contains $1 \ mole$ of solute in $1 \ L$ of the solution.

(C) The dilution formula is given by $N_1V_1 = N_2V_2$.
Here,$N_1 = \frac{1}{2} \, N$ (semi-normal),$V_1 = 200 \, cc$,and $N_2 = \frac{1}{10} \, N$ (deci-normal).
Substituting the values: $\frac{1}{2} \times 200 = \frac{1}{10} \times V_2$.
$V_2 = 1000 \, cc$.
The volume of water to be added is $V_2 - V_1 = 1000 - 200 = 800 \, cc$.

(D) Step $1$: Calculate the normality of the final solution.
Given concentration $= 10 \, mg/mL = 10 \, g/L$.
Normality $(N) = \frac{\text{Strength in } g/L}{\text{Equivalent weight}}$.
Equivalent weight of $NaOH = 40$.
$N_2 = \frac{10}{40} = 0.25 \, N$.
Step $2$: Use the dilution formula $N_1 V_1 = N_2 V_2$.
$0.5 \, N \times 500 \, mL = 0.25 \, N \times V_2$.
$V_2 = \frac{0.5 \times 500}{0.25} = 1000 \, mL$.
Step $3$: Calculate the volume of water to be added.
Volume of water added $= V_2 - V_1 = 1000 \, mL - 500 \, mL = 500 \, mL$.

(C) The number of gram molecules (moles) of a solute present in $1 \ L$ (unit volume) of a solution is defined as the molar concentration or molarity of the solution.
Mathematically,$Molarity (M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}$.

(D) The density of pure water at $4\,^{\circ}C$ is $1\,g\,mL^{-1}$.
To calculate molarity, we consider $1000\,mL$ of water.
The mass of $1000\,mL$ of water is $1000\,g$.
The molar mass of water $(H_2O)$ is $18\,g\,mol^{-1}$.
Molarity $= \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{1000\,g / 18\,g\,mol^{-1}}{1\,L} = 55.5\,M$.

(A) The formula for mixing two solutions of the same nature is $N_1V_1 + N_2V_2 = N_3V_3$.
Here,$N_1 = 0.6 \, N$,$V_1 = 200 \, mL$,$N_2 = 0.3 \, N$,and $V_2 = 100 \, mL$.
The total volume of the resultant solution is $V_3 = V_1 + V_2 = 200 \, mL + 100 \, mL = 300 \, mL$.
Substituting the values into the formula: $(0.6 \times 200) + (0.3 \times 100) = N_3 \times 300$.
$120 + 30 = N_3 \times 300$.
$150 = N_3 \times 300$.
$N_3 = \frac{150}{300} = 0.5 \, N$.

(A) The strength $(S)$ of a solution in $g/L$ is defined as the mass of the solute in grams present in $1 \ L$ of the solution.
Normality $(N)$ is defined as the number of gram equivalents of solute per liter of solution.
$N = \frac{\text{Number of gram equivalents}}{V \text{ (in } L)} = \frac{W / E}{V \text{ (in } L)}$
Where $W$ is the mass of the solute in grams and $E$ is the equivalent weight.
Rearranging the formula: $\frac{W}{V} = N \times E$
Since $\text{Strength } (S) = \frac{W}{V}$,we get $S = N \times E$.

(A) Using the dilution formula $C_1V_1 = C_2V_2$:
Given $C_1 = 40 \, mg/mL$,$C_2 = 16 \, mg/mL$,and $V_1 = 1 \, mL$.
$40 \, mg/mL \times 1 \, mL = 16 \, mg/mL \times V_2$
$V_2 = \frac{40}{16} = 2.5 \, mL$.
Therefore,each $1 \, mL$ of the original solution must be diluted to a total volume of $2.5 \, mL$.

(C) When equal volumes of two solutions are mixed,the total volume becomes double the initial volume of each solution.
Let the volume of each solution be $V$.
Total volume of the mixture = $V + V = 2V$.
The concentration of $AgNO_3$ is $0.1 \, M$. Since $AgNO_3$ dissociates completely as $AgNO_3 \rightarrow Ag^+ + NO_3^-$,the concentration of $NO_3^-$ ions in the initial solution is also $0.1 \, M$.
Using the dilution formula $M_1V_1 = M_2V_2$:
$0.1 \times V = M_2 \times 2V$
$M_2 = \frac{0.1 \times V}{2V} = 0.05 \, M$.

(B) Using the dilution formula: $N_1 V_1 = N_2 V_2$
Given: $N_1 = 4 \, N$,$V_1 = 50 \, mL$,$N_2 = 1 \, N$
Substituting the values: $4 \times 50 = 1 \times V_2$
$V_2 = 200 \, mL$
The total volume of the final solution is $200 \, mL$.
Volume of water to be added = $V_2 - V_1 = 200 \, mL - 50 \, mL = 150 \, mL$.

(A) To find the volume,we use the dilution formula: $M_1V_1 = M_2V_2$.
Here,$M_1 = 12 \; M$,$M_2 = 18 \; M$,and $V_2 = 240 \; mL$.
Substituting the values: $12 \times V_1 = 18 \times 240$.
$V_1 = \frac{18 \times 240}{12} = 360 \; mL$.
Converting to litres: $360 \; mL = 0.36 \; litre$.
Therefore,the correct option is $A$.

(C) The molarity of the final mixture is calculated using the formula: $M_1V_1 + M_2V_2 = M_{final}V_{final}$.
Given: $M_1 = 3.0 \ M$,$V_1 = 25 \ mL$,$M_2 = 4.0 \ M$,$V_2 = 75 \ mL$.
Total volume $V_{final} = V_1 + V_2 = 25 \ mL + 75 \ mL = 100 \ mL$.
Substituting the values: $(3.0 \times 25) + (4.0 \times 75) = M_{final} \times 100$.
$75 + 300 = M_{final} \times 100$.
$375 = M_{final} \times 100$.
$M_{final} = \frac{375}{100} = 3.75 \ M$.

(B) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Formula: $M = \frac{n}{V(L)}$
Given: $n = 0.006 \ mol$,$V = 100 \ mL = 0.1 \ L$
Calculation: $M = \frac{0.006 \ mol}{0.1 \ L} = 0.06 \ M$
Therefore,the correct option is $B$.

(B) The molar mass of $H_2SO_4$ is $98 \ g/mol$.
Given mass of $H_2SO_4 = 9.8 \ g$.
Volume of solution = $2 \ L$.
Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Number of moles of $H_2SO_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{9.8 \ g}{98 \ g/mol} = 0.1 \ mol$.
Molarity $(M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} = \frac{0.1 \ mol}{2 \ L} = 0.05 \ M$.

(A) The molar mass of sodium hydroxide $(NaOH)$ is $23 + 16 + 1 = 40 \ g/mol$.
The formula for molarity $(M)$ is $M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \times \frac{1000}{\text{volume of solution (mL)}}$.
Substituting the given values: $M = \frac{5}{40} \times \frac{1000}{250}$.
$M = 0.125 \times 4 = 0.5 \ M$.

(B) Molarity is defined as the number of moles of solute dissolved per litre of solution.
Molarity = $\frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$
Therefore,the unit of Molarity is $moles/litre$ (or $M$ or $mol \ L^{-1}$).
Molarity is temperature-dependent because volume changes with temperature,whereas Molality is temperature-independent.

(A) The formula for molarity $(M)$ is given by: $M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}}$.
Given: mass of $NaCl = 5.85 \ g$,molar mass of $NaCl = 58.5 \ g/mol$,volume of solution = $0.5 \ L$.
Substituting the values: $M = \frac{5.85}{58.5 \times 0.5} = \frac{0.1}{0.5} = 0.2 \ M$.
Therefore,the correct option is $(A)$.

(B) The molarity of the resultant solution is calculated using the formula $M_3 = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$.
Given: $M_1 = 1 \ M$,$V_1 = 2.5 \ L$,$M_2 = 0.5 \ M$,$V_2 = 3 \ L$.
Total moles of $NaOH = (1 \times 2.5) + (0.5 \times 3) = 2.5 + 1.5 = 4 \ mol$.
Total volume of the solution $= 2.5 \ L + 3 \ L = 5.5 \ L$.
Resultant molarity $M_3 = \frac{4 \ mol}{5.5 \ L} \approx 0.73 \ M$.

(D) When the solute is present in trace quantities,it is convenient to express the concentration in parts per million $(ppm)$.
It is defined as the mass of the solute in grams present in $10^6 \, g$ of the solution.

(B) When the concentration is expressed as the number of moles of a solute per litre of solution,it is known as $Molarity$.
The formula for $Molarity$ $(M)$ is given by:
$M = \frac{\text{number of moles of solute}}{\text{Volume of solution in litres (L)}}$

(C) The molar mass of cane sugar $(C_{12}H_{22}O_{11})$ is $12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342 \ g/mol$.
Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Number of moles of cane sugar = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{171 \ g}{342 \ g/mol} = 0.5 \ mol$.
Molarity = $\frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{0.5 \ mol}{1 \ L} = 0.5 \ M$.

(C) Let $x$ be the volume of $4\,N\, HCl$ required in liters.
Then the volume of $10\,N\, HCl$ required is $(1 - x)\, L$.
Using the mixture formula $N_1V_1 + N_2V_2 = N_3V_3$:
$4x + 10(1 - x) = 6 \times 1$
$4x + 10 - 10x = 6$
$-6x = -4$
$x = \frac{4}{6} = 0.67\, L$
Thus,the volume of $4\,N\, HCl$ is $0.67\, L$ and the volume of $10\,N\, HCl$ is $1 - 0.67 = 0.33\, L$.

(C) Concentration terms that involve volume are temperature-dependent because volume changes with temperature. $Molarity$ $(M)$ is defined as the number of moles of solute per liter of solution. Since volume of the solution changes with temperature,$Molarity$ also changes. Other terms like $Molality$,$Weight \ fraction$,and $Mole \ fraction$ are based on mass,which remains constant with temperature changes.

(C) The formula for Molarity $(M)$ is given by: $M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in Litres}}$.
Given: $M = 2.0 \, M$ and $\text{moles of solute} = 0.5 \, mol$.
Substituting the values: $2.0 = \frac{0.5}{\text{Volume of solution in L}}$.
Therefore,$\text{Volume of solution in L} = \frac{0.5}{2.0} = 0.250 \, L$.
Converting to milliliters: $0.250 \, L \times 1000 \, mL/L = 250 \, mL$ of solution.

(D) Step $1$: Calculate the number of moles of water $(n_{H_2O})$.
$n_{H_2O} = \frac{\text{mass}}{\text{molar mass}} = \frac{36 \ g}{18 \ g/mol} = 2 \ mol$.
Step $2$: Calculate the number of moles of ethyl alcohol $(n_{C_2H_5OH})$.
$n_{C_2H_5OH} = \frac{828 \ g}{46 \ g/mol} = 18 \ mol$.
Step $3$: Calculate the mole fraction of water $(x_{H_2O})$.
$x_{H_2O} = \frac{n_{H_2O}}{n_{H_2O} + n_{C_2H_5OH}} = \frac{2}{2 + 18} = \frac{2}{20} = 0.1$.

(B) Molality is defined as the number of moles of solute per $1 \ kg$ of solvent.
Since mass does not change with temperature,molality is independent of temperature.
In contrast,molarity,formality,and normality involve volume,which changes with temperature.

(B) Molality $(m)$ is defined as the number of moles of solute present in $1000 \, g$ $(1 \, kg)$ of the solvent.
Mathematically,$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
Therefore,the correct definition is the number of moles of solute per $1000 \, g$ of the solvent.

(B) The mole fraction of a component is defined as the ratio of the number of moles of that component to the total number of moles of all components in the solution.
Mole fraction of solute $= \frac{\text{moles of solute}}{\text{total moles of solution}}$
Mole fraction of solute $= \frac{20}{80} = 0.25$.

(C) The formula for normality $(N)$ is given by: $N = \frac{\text{mass in } g}{\text{Equivalent weight} \times \text{Volume in } L}$.
First,calculate the equivalent weight of $NaOH$: $\text{Eq. wt.} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{40}{1} = 40 \ g/eq$.
Given: $\text{mass} = 4 \ g$,$\text{Volume} = 100 \ mL = 0.1 \ L$.
Substituting the values: $N = \frac{4}{40 \times 0.1} = \frac{4}{4} = 1.0 \ N$.

(C) The molarity of the final mixture is calculated using the formula: $M_1V_1 + M_2V_2 = M_3V_3$
Given:
$M_1 = 1.5 \ M, V_1 = 480 \ mL$
$M_2 = 1.2 \ M, V_2 = 520 \ mL$
$V_3 = V_1 + V_2 = 480 \ mL + 520 \ mL = 1000 \ mL$
Substituting the values:
$M_3 = \frac{M_1V_1 + M_2V_2}{V_3}$
$M_3 = \frac{(1.5 \times 480) + (1.2 \times 520)}{1000}$
$M_3 = \frac{720 + 624}{1000} = \frac{1344}{1000} = 1.344 \ M$
Rounding to two decimal places,the molarity is $1.34 \ M$.

(C) Molarity is defined as the number of moles of solute dissolved per liter of solution. $\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}}$.
Therefore,a molar solution means $1 \text{ mole}$ of solute is present in $1 \text{ L}$ of solution.

(C) The formula for molality $(m)$ is given by: $m = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}}$.
Given:
Mass of glucose $= 18 \ g$
Molar mass of glucose $= 180 \ g/mol$
Mass of water (solvent) $= 500 \ g = 0.5 \ kg$.
Substituting the values:
$m = \frac{18}{180 \times 0.5} = \frac{0.1}{0.5} = 0.2 \ m$.
Therefore,the correct option is $(C)$.

(A) Given: Density $(d)$ = $1.253 \ g/mL$,Weight percentage = $22\%$,Molar mass of $Al_2(SO_4)_3$ = $342 \ g/mol$.
$1$. Molarity $(M)$ = $\frac{\% \times 10 \times d}{\text{Molar mass}} = \frac{22 \times 10 \times 1.253}{342} = 0.805 \ M$.
$2$. Normality $(N)$ = Molarity $\times$ n-factor. For $Al_2(SO_4)_3$,the n-factor is $6$ (total positive charge = $2 \times 3 = 6$).
$N = 0.805 \times 6 = 4.83 \ N$.
$3$. Molality $(m)$ = $\frac{\text{mass of solute (g)} \times 1000}{\text{Molar mass} \times \text{mass of solvent (g)}} = \frac{22 \times 1000}{342 \times (100 - 22)} = \frac{22000}{342 \times 78} = 0.825 \ m$.

(C) To find the highest normality,we calculate the normality $(N)$ for each option:
$(A)$ $8 \, g$ of $KOH / \text{litre}$: Molar mass of $KOH = 56 \, g/mol$. $N = \frac{8}{56} \times 1 = 0.143 \, N$.
$(B)$ $1 \, N$ phosphoric acid: Given as $1 \, N$.
$(C)$ $6 \, g$ of $NaOH / 100 \, mL$: Molar mass of $NaOH = 40 \, g/mol$. $N = \frac{6 \times 1000}{40 \times 100} = 1.5 \, N$.
$(D)$ $0.5 \, M \, H_2SO_4$: For $H_2SO_4$,$n$-factor = $2$. $N = M \times n\text{-factor} = 0.5 \times 2 = 1.0 \, N$.
Comparing the values: $0.143 \, N < 1.0 \, N < 1.0 \, N < 1.5 \, N$.
Thus,the solution with $6 \, g$ of $NaOH / 100 \, mL$ has the highest normality.

(C) Given: Density $(d)$ = $1.98 \ g/mL$,Percentage by weight = $98\%$,Molar mass of $H_2SO_4$ = $98 \ g/mol$,Equivalent weight of $H_2SO_4$ = $\frac{98}{2} = 49 \ g/eq$.
First,calculate the strength $(S)$ in $g/L$: $S = \text{Density} \times 1000 \times \% \text{by weight} = 1.98 \times 1000 \times 0.98 = 1940.4 \ g/L$.
Alternatively,$S = 1.98 \times 1000 \times \frac{98}{100} = 1940.4 \ g/L$.
Normality $(N)$ = $\frac{\text{Strength in } g/L}{\text{Equivalent weight}} = \frac{1940.4}{49} = 39.6 \ N$.

(C) $1 \ m$ solution means $1 \ mole$ of solute is dissolved in $1000 \ g$ of solvent $(H_2O)$.
The number of moles of water $(N)$ is calculated as: $N = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.
The mole fraction of the solute $(x_{solute})$ is given by the formula: $x_{solute} = \frac{n}{n + N}$.
Substituting the values: $x_{solute} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \approx 0.0177 \approx 0.018$.

(B) Step $1$: Calculate the initial molarity of the solution.
Initial Molarity $(M_1) = \frac{\text{mass}}{\text{molecular weight}} \times \frac{1000}{V \, (mL)}$
$M_1 = \frac{10.6}{106} \times \frac{1000}{100} = 0.1 \times 10 = 1 \, M$
Step $2$: Use the dilution formula $M_1 V_1 = M_2 V_2$ for the dilution process.
Here,$M_1 = 1 \, M$,$V_1 = 10 \, mL$,and $V_2 = 1000 \, mL$.
$1 \times 10 = M_2 \times 1000$
$M_2 = \frac{10}{1000} = 0.01 \, M = 10^{-2} \, M$
Therefore,the correct option is $B$.

(B) $20\%$ aqueous solution of $H_2O_2$ by mass means $20 \ g$ of $H_2O_2$ is present in $100 \ g$ of the solution.
Therefore,the mass of water = $100 \ g - 20 \ g = 80 \ g$.
Moles of $H_2O_2$ $(n_{H_2O_2})$ = $\frac{20 \ g}{34 \ g/mol} = \frac{20}{34} \ mol$.
Moles of $H_2O$ $(n_{H_2O})$ = $\frac{80 \ g}{18 \ g/mol} = \frac{80}{18} \ mol$.
Mole fraction of water $(x_{H_2O})$ = $\frac{n_{H_2O}}{n_{H_2O} + n_{H_2O_2}} = \frac{\frac{80}{18}}{\frac{80}{18} + \frac{20}{34}}$.
$x_{H_2O} = \frac{4.44}{4.44 + 0.588} = \frac{4.44}{5.028} \approx 0.883$.
Calculating the fraction: $\frac{80/18}{(80 \times 34 + 20 \times 18) / (18 \times 34)} = \frac{80 \times 34}{2720 + 360} = \frac{2720}{3080} = \frac{272}{308} = \frac{68}{77}$.

(D) Mole fraction is defined as the ratio of the number of moles of a specific constituent to the total number of moles of all constituents present in the solution.
Mathematically,for a component $i$,the mole fraction $X_i$ is given by:
$X_i = \frac{n_i}{n_{total}} = \frac{\text{No. of moles of any constituent}}{\text{Total no. of moles of all constituents}}$

(B) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Number of moles of solute $= \frac{W_B}{M_B}$.
Mass of solvent in kg $= \frac{W_A}{1000}$.
Therefore,$m = \frac{W_B / M_B}{W_A / 1000} = \frac{W_B}{M_B} \times \frac{1000}{W_A}$.

(B) The normality of a solution is defined as the number of gram equivalents of a solute present per liter of solution. It is denoted by the symbol $N$.
$N = \frac{\text{Number of gram equivalents of solute}}{\text{Volume of solution in litres}}$
Mathematically,it can be expressed as:
$N = \frac{\text{Weight of solute}}{\text{Equivalent weight of solute}} \times \frac{1}{\text{Volume of solution in litres}}$
Where,$\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}}$.

(A) The concentration of the solution is given as $0.03 \ g/mL$.
This means that $1 \ mL$ of solution contains $0.03 \ g$ of $AgNO_3$.
To find the amount of $AgNO_3$ in $60 \ mL$ of solution,we multiply the concentration by the total volume:
$\text{Amount of } AgNO_3 = 60 \ mL \times 0.03 \ g/mL = 1.8 \ g$.

(A) The molar mass of $Na_2CO_3$ is $(2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$.
The formula for molarity $(M)$ is $M = \frac{w \times 1000}{m.wt. \times V \text{ (in mL)}}$.
Substituting the given values: $M = \frac{10.6 \times 1000}{106 \times 500} = \frac{10600}{53000} = 0.2 \ M$.

(D) Molality is defined as the number of moles of solute present in $1 \ kg$ of solvent.
The formula is: $\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.

(C) The molarity $(M)$ of the solution is calculated as: $M = \frac{\text{mass of solute}}{\text{molar mass} \times \text{volume of solution in litres}} = \frac{18 \ g}{180 \ g/mol \times 1 \ L} = 0.1 \ M$.
Assuming the density of the solution is approximately $1 \ g/mL$,the mass of $1 \ L$ of solution is $1000 \ g$.
The mass of the solvent (water) is $1000 \ g - 18 \ g = 982 \ g = 0.982 \ kg$.
Molality $(m)$ is defined as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \ mol}{0.982 \ kg} \approx 0.102 \ molal$.
Given the options provided and the context of such problems,the intended answer is $0.1$.