Method of expressing concentration of solution Questions in English

(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given $m = 5.2 \ mol/kg$,this means there are $5.2 \ moles$ of $CH_3OH$ in $1000 \ g$ $(1 \ kg)$ of water.
Moles of water $(n_{H_2O})$ = $\frac{1000 \ g}{18.02 \ g/mol} \approx 55.55 \ moles$.
Mole fraction of $CH_3OH$ $(x_{CH_3OH})$ = $\frac{n_{CH_3OH}}{n_{CH_3OH} + n_{H_2O}}$.
$x_{CH_3OH} = \frac{5.2}{5.2 + 55.55} = \frac{5.2}{60.75} \approx 0.0856$.
Rounding to the nearest provided option,the value is approximately $0.08$.

(A) Given: Mole fraction of solute $(x_2)$ = $0.2$.
Therefore,mole fraction of solvent (benzene,$x_1$) = $1 - 0.2 = 0.8$.
Molality $(m)$ is given by the formula: $m = \frac{x_2 \times 1000}{x_1 \times M_1}$,where $M_1$ is the molar mass of the solvent (benzene,$C_6H_6$).
Molar mass of benzene $(M_1)$ = $(6 \times 12) + (6 \times 1) = 78 \ g/mol$.
Substituting the values: $m = \frac{0.2 \times 1000}{0.8 \times 78} = \frac{200}{62.4} \approx 3.205 \ m$.
Rounding to the nearest option,$X \approx 3.2$.

(A) For $HCl$,the normality $(N)$ is equal to the molarity $(M)$ because the $n$-factor is $1$.
Thus,$0.5 \ (N) \ HCl = 0.5 \ (M) \ HCl$.
The formula for the molarity of a mixture is $M_{mix} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$.
Here,$M_1 = 0.5 \ M$,$V_1 = 750 \ mL$,$M_2 = 2 \ M$,and $V_2 = 250 \ mL$.
$M_{mix} = \frac{(0.5 \times 750) + (2 \times 250)}{750 + 250} = \frac{375 + 500}{1000} = \frac{875}{1000} = 0.875 \ M$.
Rounding to two decimal places,the answer is $0.87 \ M$.

(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Statement-$2$ correctly defines $1$ molal solution as $1$ mole of solute in $1000 \ g$ $(1 \ kg)$ of water.
For glucose $(C_6H_{12}O_6)$,the molar mass is $180 \ g/mol$.
Therefore,$1$ mole of glucose is $180 \ g$.
Since Statement-$1$ states that $1$ molal glucose solution contains $180 \ g$ of glucose in $1000 \ g$ of water,it is also true.
Statement-$2$ provides the definition of molality,which explains why $1$ mole $(180 \ g)$ of glucose in $1000 \ g$ of water constitutes a $1$ molal solution.
Thus,Statement-$2$ is the correct explanation for Statement-$1$.

(B) Hardness is given as $50 \, ppm$,which means $50 \, mg$ of solute per $10^6 \, mg$ of water.
$10^6 \, mg$ of water is equal to $10^3 \, g$ or $1 \, kg$ of water.
Since the hardness is expressed in terms of $MgSO_4$ concentration directly in this context,$50 \, ppm$ of $MgSO_4$ means $50 \, mg$ of $MgSO_4$ per $10^6 \, mg$ of water.
Therefore,in $1 \, kg$ of water,there are $50 \, mg$ of $MgSO_4$.

(C) The concentration in $ppm$ is calculated using the formula:
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$
Given:
Mass of solute $(F^-)$ = $0.2 \ g$
Mass of solution = $500 \ g$
$ppm = \frac{0.2}{500} \times 10^6 = 0.0004 \times 10^6 = 400 \ ppm$
Therefore,the correct option is $C$.

(B) The number of moles of $NaCl$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.85 \, g}{58.5 \, g/mol} = 0.1 \, mol$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{n_{solute}}{W_{solvent(kg)}} = \frac{0.1 \, mol}{1 \, kg} = 0.1 \, mol/kg$.
Therefore,the strength of the solution is $0.1 \, m$ (molal).

(B) Step $1$: Calculate the molarity of $Na_2CO_3$ solution.
Molarity $(M) = \frac{\text{mass of solute}}{\text{molar mass} \times \text{volume of solution in } L}$
$M = \frac{25.3 \ g}{106 \ g \ mol^{-1} \times 0.250 \ L} = 0.955 \ M$
Step $2$: Write the dissociation equation.
$Na_2CO_3(aq) \rightarrow 2Na^{+}(aq) + CO_3^{2-}(aq)$
Step $3$: Determine the concentration of ions.
Since $1 \ mol$ of $Na_2CO_3$ produces $2 \ mol$ of $Na^{+}$ and $1 \ mol$ of $CO_3^{2-}$,
$[Na^{+}] = 2 \times 0.955 \ M = 1.910 \ M$
$[CO_3^{2-}] = 1 \times 0.955 \ M = 0.955 \ M$
Thus,the concentrations are $1.910 \ M$ and $0.955 \ M$ respectively.

(A) Molarity is defined as the number of moles of solute dissolved per liter of solution $(M = \frac{n}{V(L)})$.
Since the volume of a solution changes with temperature due to thermal expansion or contraction,molarity is temperature-dependent.
In contrast,mole fraction,weight percentage,and molality are based on mass,which remains constant regardless of temperature changes.

(C) $1.00 \, m$ solution means $1 \, mol$ of solute is present in $1000 \, g$ of water.
The number of moles of water is $n_{H_{2}O} = \frac{1000 \, g}{18 \, g/mol} = 55.5 \, mol$.
The mole fraction of the solute is given by $X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{H_{2}O}}$.
Substituting the values,$X_{\text{solute}} = \frac{1}{1 + 55.5} = \frac{1}{56.5} \approx 0.0177$.

(A) Given: Molarity $(M) = 2.05 \ M$,Density $(d) = 1.02 \ g/mL$,Molar mass of acetic acid $(CH_3COOH) = 60 \ g/mol$.
Molality $(m)$ is calculated using the formula: $m = \frac{1000 \times M}{1000 \times d - M \times M_{solute}}$.
Substituting the values: $m = \frac{1000 \times 2.05}{1000 \times 1.02 - 2.05 \times 60}$.
$m = \frac{2050}{1020 - 123} = \frac{2050}{897} \approx 2.28 \ mol \ kg^{-1}$.

(C) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given $m = 5.2 \ mol/kg$,this means $5.2$ moles of $CH_3OH$ are dissolved in $1000 \ g$ of water $(H_2O)$.
Moles of water $(n_{H_2O})$ $= \frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Mole fraction of methyl alcohol $(X_{CH_3OH})$ $= \frac{n_{CH_3OH}}{n_{CH_3OH} + n_{H_2O}}$.
$X_{CH_3OH} = \frac{5.2}{5.2 + 55.56} = \frac{5.2}{60.76} \approx 0.0856$.
Rounding to two significant figures,we get $0.086$.

(D) $1$. Calculate the number of moles of solute (urea): $\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{120 \ g}{60 \ g/mol} = 2 \ mol$.
$2$. Calculate the total mass of the solution: $\text{Mass of solution} = \text{mass of solute} + \text{mass of solvent} = 120 \ g + 1000 \ g = 1120 \ g$.
$3$. Calculate the volume of the solution using density: $\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{1120 \ g}{1.15 \ g/mL} \approx 973.91 \ mL = 0.97391 \ L$.
$4$. Calculate the molarity $(M)$: $M = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{2 \ mol}{0.97391 \ L} \approx 2.05 \ M$.

(A) The molarity of the final mixture is calculated using the formula:
$M_{mix} = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$
Given:
$M_{1} = 0.5 \, M$,$V_{1} = 750 \, mL$
$M_{2} = 2 \, M$,$V_{2} = 250 \, mL$
Total volume $V = 750 \, mL + 250 \, mL = 1000 \, mL$
Substituting the values:
$M_{mix} = \frac{(0.5 \times 750) + (2 \times 250)}{1000}$
$M_{mix} = \frac{375 + 500}{1000} = \frac{875}{1000} = 0.875 \, M$

(C) Let the volume of each liquid be $V = 100 \ mL$.
Mass of $A = d_A \times V = 0.8 \ g/mL \times 100 \ mL = 80 \ g$.
Mass of $B = d_B \times V = 1.2 \ g/mL \times 100 \ mL = 120 \ g$.
Moles of $A$ $(n_A)$ = $80 / 16 = 5 \ mol$.
Moles of $B$ $(n_B)$ = $120 / 32 = 3.75 \ mol$.
Mole fraction of $A$ $(X_A)$ = $n_A / (n_A + n_B) = 5 / (5 + 3.75) = 5 / 8.75 = 500 / 875 = 4/7$.

(A) Given: Molality $(m)$ = $3.0 \ mol/kg$,Density $(d)$ = $1.110 \ g/mL$,Molar mass of $NaOH$ $(M_B)$ = $40 \ g/mol$.
Let the mass of the solvent be $1 \ kg = 1000 \ g$.
The mass of the solution = Mass of solute + Mass of solvent = $(3.0 \times 40) + 1000 = 120 + 1000 = 1120 \ g$.
The volume of the solution = $\frac{\text{Mass of solution}}{\text{Density}} = \frac{1120 \ g}{1.110 \ g/mL} \approx 1009.01 \ mL = 1.009 \ L$.
Molarity $(M)$ = $\frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{3.0 \ mol}{1.009 \ L} \approx 2.97 \ M$.
Using the formula $M = \frac{m \times d \times 1000}{1000 + (m \times M_B)} = \frac{3.0 \times 1.110 \times 1000}{1000 + (3.0 \times 40)} = \frac{3330}{1120} \approx 2.97 \ M$.
The closest option is $2.94$.

(B) Mass of solution $= 100 \ g$.
Mass of $H_2SO_4 = 13 \ g$.
Volume of solution $= \frac{100 \ g}{1.09 \ g/mL} = 91.74 \ mL = 0.09174 \ L$.
Moles of $H_2SO_4 = \frac{13 \ g}{98 \ g/mol} = 0.1326 \ mol$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.1326}{0.09174} \approx 1.445 \ M$.
Normality $(N) = \text{Molarity} \times n\text{-factor}$.
For $H_2SO_4$,$n\text{-factor} = 2$.
Normality $= 1.445 \times 2 = 2.89 \ N$.

(B) $1$. Calculate moles of $Cl^-$ from $KCl$: $n(Cl^-) = M \times V(L) = 0.5 \times 0.2 = 0.1 \ mol$.
$2$. Calculate moles of $Cl^-$ from $MgCl_2$: $19 \% \ w/v$ means $19 \ g$ in $100 \ mL$,so $50 \ mL$ contains $9.5 \ g$ of $MgCl_2$.
Molar mass of $MgCl_2 = 24.3 + 2 \times 35.5 = 95.3 \ g/mol \approx 95 \ g/mol$.
Moles of $MgCl_2 = 9.5 / 95 = 0.1 \ mol$.
Since $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$,$n(Cl^-) = 2 \times 0.1 = 0.2 \ mol$.
$3$. Total moles of $Cl^- = 0.1 + 0.2 = 0.3 \ mol$.
$4$. Total volume before dilution = $200 \ mL + 50 \ mL = 250 \ mL = 0.25 \ L$.
$5$. Final volume after $8$ times dilution = $0.25 \ L \times 8 = 2 \ L$.
$6$. Molarity of $Cl^-$ = $\text{Total moles} / \text{Final volume} = 0.3 \ mol / 2 \ L = 0.15 \ M$.

(D) Case $-I$: $(V_1 : V_2 = x : y)$
$M_{res} = 1.5 = \frac{3x + 1y}{x + y}$
$1.5x + 1.5y = 3x + y$
$0.5y = 1.5x \implies y = 3x$
Case $-II$: $(V_1 : V_2 = y : x)$
$M_{res} = \frac{3y + 1x}{y + x}$
Substituting $y = 3x$:
$M_{res} = \frac{3(3x) + x}{3x + x} = \frac{9x + x}{4x} = \frac{10x}{4x} = 2.5 \, M$.

(A) Given: Molarity $(M)$ = $2.55 \, M$,Mass percentage = $20\% \,(W/W)$,Molar mass of $H_2SO_4$ = $98 \, g \, mol^{-1}$.
Let the mass of the solution be $100 \, g$.
Mass of solute $(H_2SO_4)$ = $20 \, g$.
Number of moles of solute $(n)$ = $\frac{20}{98} \, mol$.
Volume of solution $(V)$ = $\frac{\text{Mass}}{\text{Density}} = \frac{100}{d} \, mL = \frac{100}{1000d} \, L = \frac{1}{10d} \, L$.
Molarity $(M)$ = $\frac{n}{V(L)} = \frac{20/98}{1/10d} = \frac{200d}{98}$.
Given $M = 2.55$,so $2.55 = \frac{200d}{98}$.
$d = \frac{2.55 \times 98}{200} = 1.2495 \approx 1.25 \, g \, cm^{-3}$.

(NONE) Total volume $= 500 \ mL + 1500 \ mL = 2000 \ mL = 2 \ L$.
Moles of $Na^{+}$ ions $= 0.2 \ M \times 0.5 \ L = 0.1 \ mol$.
Concentration of $Na^{+}$ ions $= \frac{0.1 \ mol}{2 \ L} = 0.05 \ M$.
Moles of $Mg^{2+}$ ions $= 0.4 \ M \times 1.5 \ L = 0.6 \ mol$.
Concentration of $Mg^{2+}$ ions $= \frac{0.6 \ mol}{2 \ L} = 0.3 \ M$.
Mass concentration of $Mg^{2+}$ $= 0.3 \ mol/L \times 24 \ g/mol = 7.2 \ g/L$.
Moles of $Cl^{-}$ ions $= (0.2 \times 0.5) + (2 \times 0.4 \times 1.5) = 0.1 + 1.2 = 1.3 \ mol$.
Concentration of $Cl^{-}$ ions $= \frac{1.3 \ mol}{2 \ L} = 0.65 \ M$.
Since all calculated values match the options,there is no incorrect statement provided in the options.

(D) Given: $15\% \,(w/V)$ solution of $H_2SO_4$ means $15\, g$ of $H_2SO_4$ is present in $100\, mL$ of solution.
Density of solution = $1.1\, g/mL$.
Mass of solution = $\text{Volume} \times \text{Density} = 100\, mL \times 1.1\, g/mL = 110\, g$.
Mass of solvent = $\text{Mass of solution} - \text{Mass of solute} = 110\, g - 15\, g = 95\, g = 0.095\, kg$.
Molar mass of $H_2SO_4 = 98\, g/mol$.
Moles of solute = $\frac{15\, g}{98\, g/mol} \approx 0.153\, mol$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{0.153\, mol}{0.095\, kg} \approx 1.61\, mol/kg$.
Thus,the correct option is $D$.

(A) Let the volume of solution $A$ be $V_1$ and the volume of solution $B$ be $V_2$.
Given $V_1 + V_2 = 2 \, L$,so $V_2 = (2 - V_1) \, L$.
Using the molarity equation $M_1 V_1 + M_2 V_2 = M_3 V_3$:
$0.5 \times V_1 + 0.1 \times (2 - V_1) = 0.2 \times 2$
$0.5 V_1 + 0.2 - 0.1 V_1 = 0.4$
$0.4 V_1 = 0.2$
$V_1 = \frac{0.2}{0.4} = 0.5 \, L$
Therefore,$V_2 = 2 - 0.5 = 1.5 \, L$.
Thus,$0.5 \, L$ of $A$ and $1.5 \, L$ of $B$ are required.

(C) First,calculate the number of moles of $NaOH$ and $H_2O$:
$n_{NaOH} = \frac{80 \ g}{40 \ g/mol} = 2 \ mol$
$n_{H_2O} = \frac{54 \ g}{18 \ g/mol} = 3 \ mol$
The mole fraction of water $(X_{H_2O})$ is given by:
$X_{H_2O} = \frac{n_{H_2O}}{n_{H_2O} + n_{NaOH}}$
$X_{H_2O} = \frac{3}{3 + 2} = \frac{3}{5} = 0.6$

(C) Step $1$: Calculate the moles of solute $(H_2SO_4)$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{98 \times 10^{-3} \, g}{98 \, g/mol} = 10^{-3} \, mol$.
Step $2$: Calculate the volume of the solution:
$V = \frac{\text{mass of solution}}{\text{density}} = \frac{100 \, g}{1.25 \, g/mL} = 80 \, mL = 0.08 \, L$.
Step $3$: Calculate molarity $(M)$:
$M = \frac{n}{V(L)} = \frac{10^{-3} \, mol}{0.08 \, L} = 0.0125 \, M = 1.25 \times 10^{-2} \, M$.

(C) The relation between molarity $(M)$ and molality $(m)$ is given by the formula: $d = M \left( \frac{1}{m} + \frac{M_2}{1000} \right)$,where $d$ is density in $\text{g mL}^{-1}$,$M$ is molarity,$m$ is molality,and $M_2$ is the molar mass of the solute.
Given: $M = 3 \text{ M}$,$d = 1.252 \text{ g mL}^{-1}$,$M_2 = 58.5 \text{ g mol}^{-1}$.
Substituting the values: $1.252 = 3 \left( \frac{1}{m} + \frac{58.5}{1000} \right)$.
$1.252 / 3 = \frac{1}{m} + 0.0585$.
$0.41733 = \frac{1}{m} + 0.0585$.
$\frac{1}{m} = 0.41733 - 0.0585 = 0.35883$.
$m = 1 / 0.35883 \approx 2.79 \text{ m}$.

(A) Using the molarity equation for mixing two solutions of the same solute:
$M_1V_1 + M_2V_2 = M_{final}V_{total}$
Given:
$M_1 = 2 \, M, V_1 = 10 \, mL$
$M_2 = 0.5 \, M, V_2 = 200 \, mL$
$V_{total} = 10 \, mL + 200 \, mL = 210 \, mL$
Substituting the values:
$(2 \times 10) + (0.5 \times 200) = M_{final} \times 210$
$20 + 100 = M_{final} \times 210$
$120 = M_{final} \times 210$
$M_{final} = \frac{120}{210} = \frac{12}{21} = \frac{4}{7} \approx 0.57 \, M$

(A) The formula for parts per million $(ppm)$ is given by:
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$
Given:
Mass of solute $(F^{-})$ = $0.2 \ g$
Mass of solution = $500 \ g$
Calculation:
$ppm = \frac{0.2}{500} \times 10^6$
$ppm = 0.0004 \times 10^6$
$ppm = 400$

(B) The molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given mass of $Na^{+}$ ions = $92 \ g$.
Molar mass of $Na^{+}$ ions = $23 \ g \ mol^{-1}$.
Moles of $Na^{+}$ ions = $\frac{92 \ g}{23 \ g \ mol^{-1}} = 4 \ mol$.
Mass of solvent (water) = $1 \ kg$.
Molality = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{4 \ mol}{1 \ kg} = 4 \ mol \ kg^{-1}$.

(C) The molarity $(M)$ is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V(L)}$.
Given $M = 0.1 \ M$ and $V = 2 \ L$,the number of moles $(n)$ is $n = M \times V = 0.1 \times 2 = 0.2 \ mol$.
The molar mass of sugar $(C_{12}H_{22}O_{11})$ is $(12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \ g/mol$.
The mass required is $mass = n \times \text{molar mass} = 0.2 \ mol \times 342 \ g/mol = 68.4 \ g$.

(C) Moles of $NaOH = \frac{8 \ g}{40 \ g \ mol^{-1}} = 0.2 \ mol$.
Moles of $H_2O = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$.
Mole fraction of $NaOH = \frac{n_{NaOH}}{n_{NaOH} + n_{H_2O}} = \frac{0.2}{0.2 + 1} = \frac{0.2}{1.2} = 0.167$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.2 \ mol}{18 \times 10^{-3} \ kg} = \frac{0.2}{0.018} \approx 11.11 \ mol \ kg^{-1}$.

(C) $20\% \, w/w$ aqueous solution of $KI$ means $20 \, g$ of $KI$ is present in $100 \, g$ of solution.
Mass of solute $(KI) = 20 \, g$.
Mass of solvent (water) $= 100 \, g - 20 \, g = 80 \, g = 0.08 \, kg$.
Molar mass of $KI = 166 \, g \, mol^{-1}$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.
Moles of $KI = \frac{20 \, g}{166 \, g \, mol^{-1}} \approx 0.1205 \, mol$.
Molality $(m) = \frac{0.1205 \, mol}{0.08 \, kg} = 1.506 \, mol \, kg^{-1} \approx 1.51 \, m$.

(C) Given: Mole fraction of solvent $(X_{solvent})$ = $0.8$.
Since it is an aqueous solution,the solvent is water $(H_2O)$,with molar mass $M_{solvent} = 18 \ g \ mol^{-1}$.
Mole fraction of solute $(X_{solute})$ = $1 - X_{solvent} = 1 - 0.8 = 0.2$.
Let the total number of moles be $n_{total} = 1$. Then $n_{solute} = 0.2 \ mol$ and $n_{solvent} = 0.8 \ mol$.
Mass of solvent = $n_{solvent} \times M_{solvent} = 0.8 \ mol \times 18 \ g \ mol^{-1} = 14.4 \ g = 0.0144 \ kg$.
Molality $(m)$ = $\frac{n_{solute}}{\text{Mass of solvent in } kg} = \frac{0.2 \ mol}{0.0144 \ kg} = 13.88 \ mol \ kg^{-1}$.

(C) Using the dilution formula: $M_{1}V_{1} = M_{2}V_{2}$
Given: $M_{1} = 0.25 \ M$,$V_{1} = 40 \ cc$,$M_{2} = 0.1 \ M$.
Calculating the final volume $V_{2}$:
$V_{2} = \frac{M_{1}V_{1}}{M_{2}} = \frac{0.25 \times 40}{0.1} = 100 \ cc$.
The volume of water to be added is $V_{2} - V_{1} = 100 - 40 = 60 \ cc$.

(A) The dissociation reactions are:
$KCl \rightarrow K^{+} + Cl^{-}$
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^{-}$
For $KCl$ solution:
$Moles \ of \ Cl^{-} = M \times V = 3.0 \ M \times 0.3 \ L = 0.9 \ mol$
For $BaCl_2$ solution:
$Moles \ of \ Cl^{-} = 2 \times M \times V = 2 \times 4.0 \ M \times 0.2 \ L = 1.6 \ mol$
Total moles of $Cl^{-} = 0.9 \ mol + 1.6 \ mol = 2.5 \ mol$
Total volume of solution = $300 \ mL + 200 \ mL = 500 \ mL = 0.5 \ L$
$[Cl^{-}] = \frac{\text{Total moles of } Cl^{-}}{\text{Total volume in } L} = \frac{2.5 \ mol}{0.5 \ L} = 5.0 \ M$

(B) Let the volume of the solution be $1 \ L$ $(1000 \ mL)$.
Then,the number of moles of solute is $M$.
The mass of the solution is $d \times 1000 \ g$.
The mass of the solute is $M \times M_{solute} \ g$.
The mass of the solvent is $(1000d - MM_{solute}) \ g$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{M \times 1000}{1000d - MM_{solute}}$.
Rearranging this,we get $\frac{1}{m} = \frac{1000d - MM_{solute}}{1000M} = \frac{d}{M} - \frac{M_{solute}}{1000}$.

(B) Given: Molarity $(M)$ = $1 \ M$,Density $(d)$ = $1.0585 \ g/mL$,Molar mass of $NaCl$ = $58.5 \ g/mol$.
Consider $1 \ L$ $(1000 \ mL)$ of solution.
Moles of $NaCl$ = $1 \ mol$.
Mass of $NaCl$ = $1 \ mol \times 58.5 \ g/mol = 58.5 \ g$.
Mass of solution = $\text{Volume} \times \text{Density} = 1000 \ mL \times 1.0585 \ g/mL = 1058.5 \ g$.
Mass of solvent $(H_2O)$ = $\text{Mass of solution} - \text{Mass of solute} = 1058.5 \ g - 58.5 \ g = 1000 \ g = 1 \ kg$.
Molality $(m)$ = $\frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{1 \ mol}{1 \ kg} = 1 \ m$.

(C) The number of moles of substance in solution $A$ is $n_A = M_A \times V_A = 0.1 \ M \times 100 \ mL = 10 \ mmol$.
The number of moles of substance in solution $B$ is $n_B = M_B \times V_B = 0.2 \ M \times 25 \ mL = 5 \ mmol$.
The total number of moles is $n_{total} = 10 \ mmol + 5 \ mmol = 15 \ mmol$.
The total volume of the mixture is $V_{total} = 100 \ mL + 25 \ mL = 125 \ mL$.
The final molarity $M_{final} = \frac{n_{total}}{V_{total}} = \frac{15 \ mmol}{125 \ mL} = 0.12 \ M$.

(A) Normality $(N)$ is calculated as: $N = \frac{\text{Mass concentration (g/L)}}{\text{Equivalent mass}}$.
For $HCl$: Equivalent mass = $36.5 \ g/mol$. Concentration = $0.03659 \ g/mL = 36.59 \ g/L$.
$N_{HCl} = \frac{36.59}{36.5} \approx 1.002 \ N$.
For $CH_3COOH$: Equivalent mass = $60 \ g/mol$. Concentration = $0.04509 \ g/mL = 45.09 \ g/L$.
$N_{CH_3COOH} = \frac{45.09}{60} \approx 0.7515 \ N$.
Comparing the two,$1.002 > 0.7515$,therefore $N_{HCl}$ is more.

(C) Let the mass of solute be $W_B = 100 \ g$ and mass of solvent be $W_A = 1400 \ g$.
Total mass of solution $= W_A + W_B = 1400 + 100 = 1500 \ g$.
Density of solution $d = 1.5 \ g/mL$.
Volume of solution $V = \frac{\text{mass}}{d} = \frac{1500 \ g}{1.5 \ g/mL} = 1000 \ mL = 1 \ L$.
Molarity $(M) = \frac{n_B}{V(L)} = \frac{W_B / M_B}{1} = \frac{100}{M_B}$.
Molality $(m) = \frac{n_B}{W_A(kg)} = \frac{100 / M_B}{1400 / 1000} = \frac{100}{M_B} \times \frac{1000}{1400} = \frac{100}{M_B} \times \frac{10}{14}$.
Ratio $\frac{M}{m} = \frac{100 / M_B}{(100 / M_B) \times (10 / 14)} = \frac{14}{10} = 1.4$.

(A) $2.5 \ m$ aqueous solution means $2.5 \ mol$ of solute is present in $1 \ kg$ $(1000 \ g)$ of solvent $(water)$.
$\text{Moles of solute} = 2.5 \ mol$
$\text{Moles of water} = \frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$
$\text{Mole fraction of solute} (\chi_{solute}) = \frac{n_{solute}}{n_{solute} + n_{solvent}}$
$\chi_{solute} = \frac{2.5}{2.5 + 55.56} = \frac{2.5}{58.06} \approx 0.043$

(B) The formula for molarity $(M)$ is given by: $M = \frac{\text{mass percentage} \times \text{density} \times 10}{\text{molar mass of solute}}$.
Given:
Mass percentage = $12.25\%$
Density $(d)$ = $1.056 \ g/mL$
Molar mass of $H_2SO_4$ = $98 \ g/mol$.
Calculation:
$M = \frac{12.25 \times 1.056 \times 10}{98}$
$M = \frac{129.36}{98} = 1.32 \ M$.

(D) Given $M_1 = 3\,M$ and $M_2 = 1\,M$.
Case $1$: When mixed in the ratio $x : y$ by volume,the resultant molarity $M_R$ is given by:
$M_R = \frac{M_1x + M_2y}{x + y} = 1.5$
$\frac{3x + y}{x + y} = 1.5$
$3x + y = 1.5x + 1.5y$
$1.5x = 0.5y \Rightarrow y = 3x$.
Case $2$: When mixed in the ratio $y : x$ by volume,the new resultant molarity $M'_R$ is:
$M'_R = \frac{M_1y + M_2x}{y + x}$
Substituting $y = 3x$:
$M'_R = \frac{3(3x) + 1(x)}{3x + x} = \frac{9x + x}{4x} = \frac{10x}{4x} = 2.5\,M$.

(C) The formula for the molarity of a mixture of two solutions of the same solute is $M_{result} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$.
Given:
$M_1 = 0.5 \, M, V_1 = 20 \, mL$
$M_2 = 0.3 \, M, V_2 = 30 \, mL$
Substituting the values:
$M_{result} = \frac{(0.5 \times 20) + (0.3 \times 30)}{20 + 30}$
$M_{result} = \frac{10 + 9}{50} = \frac{19}{50} = 0.38 \, M$.

(D) Initial volume of $NaOH$ solution = $500 \ mL$ and molarity = $0.5 \ M$.
Number of moles of $NaOH = \text{Molarity} \times \text{Volume (in L)} = 0.5 \times 0.5 = 0.25 \ mol$.
Mass of $NaOH = \text{Moles} \times \text{Molar mass} = 0.25 \times 40 \ g = 10 \ g$.
Since $1 \ g = 1000 \ mg$,mass of $NaOH = 10000 \ mg$.
Let the volume of water added be $V \ mL$. The final volume of the solution = $(500 + V) \ mL$.
Final strength = $\frac{\text{Mass of solute (mg)}}{\text{Volume of solution (mL)}} = \frac{10000}{500 + V} = 10 \ mg \ mL^{-1}$.
$1000 = 500 + V$.
$V = 500 \ mL$.

(D) The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \, g/mol$.
$A$ $10 \%$ aqueous solution means $10 \, g$ of glucose is present in $100 \, mL$ of the solution.
To find the volume containing $1 \, g-mole$ $(180 \, g)$ of glucose:
Volume $= \frac{100 \, mL}{10 \, g} \times 180 \, g = 1800 \, mL$.
Converting to liters: $1800 \, mL = 1.8 \, L$.

(D) $1$. Calculate moles of solute: $n = \frac{25 \ g}{250 \ g \ mol^{-1}} = 0.1 \ mol$.
$2$. Calculate total mass of solution: $Mass_{solution} = Mass_{solute} + Mass_{solvent} = 25 \ g + 100 \ g = 125 \ g$.
$3$. Calculate volume of solution: $V = \frac{Mass_{solution}}{Density} = \frac{125 \ g}{1.25 \ g \ mL^{-1}} = 100 \ mL = 0.1 \ L$.
$4$. Calculate molarity $(M)$: $M = \frac{n}{V(L)} = \frac{0.1 \ mol}{0.1 \ L} = 1 \ M$.
$5$. Calculate molality $(m)$: $m = \frac{n}{Mass_{solvent}(kg)} = \frac{0.1 \ mol}{0.1 \ kg} = 1 \ m$.
Thus,the molarity and molality are $1 \ M$ and $1 \ m$ respectively.

(B) The formula for mass percentage is: $\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solute} + \text{Mass of solvent}} \times 100$
Let the mass of $KCl$ be $W \ g$.
Given: $\text{Mass of solvent} (H_2O) = 60 \ g$,$\text{Percentage} = 20 \%$.
Substituting the values:
$\frac{W}{W + 60} \times 100 = 20$
Divide both sides by $20$:
$\frac{W}{W + 60} \times 5 = 1$
$5W = W + 60$
$4W = 60$
$W = 15 \ g$

(D) Given: Density $(d)$ = $0.6 \, g/mL$,Mass percentage = $35 \%$,Molar mass of $NH_4OH$ = $35 \, g/mol$.
Since $NH_4OH$ is a monoacidic base,its equivalent mass $(E)$ is equal to its molar mass,$E = 35 \, g/eq$.
Consider $100 \, g$ of the solution. The mass of $NH_4OH$ $(W)$ = $35 \, g$.
The volume of the solution $(V)$ = $\frac{\text{mass}}{\text{density}} = \frac{100 \, g}{0.6 \, g/mL} = \frac{100}{0.6} \, mL = \frac{100}{0.6 \times 1000} \, L$.
Normality $(N)$ = $\frac{W}{E \times V(L)} = \frac{35}{35 \times (100 / (0.6 \times 1000))} = \frac{0.6 \times 1000}{100} = 6 \, N$.