Method of expressing concentration of solution Questions in English

(B) $10\%\ (v/v)$ solution means $10\ mL$ of solute $(Br_2)$ is present in $100\ mL$ of total solution.
Therefore,the volume of solvent $(CCl_4)$ $= 100\ mL - 10\ mL = 90\ mL$.
Mass of solute $(Br_2)$ $= \text{Volume} \times \text{Density} = 10\ mL \times 3.2\ g\ cm^{-3} = 32\ g$.
Mass of solvent $(CCl_4)$ $= \text{Volume} \times \text{Density} = 90\ mL \times 1.6\ g\ cm^{-3} = 144\ g = 0.144\ kg$.
Moles of solute $(Br_2)$ $= \frac{32\ g}{160\ g\ mol^{-1}} = 0.2\ mol$.
Molality $(m)$ $= \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.2}{0.144} \approx 1.3888\ m$.
Expressing as $x \times 10^{-2}\ m$,we get $138.88 \times 10^{-2}\ m$.
Rounding to the nearest integer,$x = 139$.

(D) Mass of sugar in the first solution $= 200 \ g \times 0.25 = 50 \ g$.
Mass of sugar in the second solution $= 500 \ g \times 0.40 = 200 \ g$.
Total mass of sugar $= 50 \ g + 200 \ g = 250 \ g$.
Total mass of the resulting solution $= 200 \ g + 500 \ g = 700 \ g$.
Mass percentage of the resulting sugar solution $= (\frac{250 \ g}{700 \ g}) \times 100 \approx 35.71 \%$.
Rounding to the nearest integer,we get $36 \%$.

(D) Mass of solute $(X) = 2 \ g$
Mass of solvent $(H_2O) = 1 \ mole = 18 \ g$
Total mass of solution = Mass of solute + Mass of solvent = $2 \ g + 18 \ g = 20 \ g$
Mass percent of $X = \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100$
Mass percent of $X = \frac{2 \ g}{20 \ g} \times 100 = 10 \ \%$

(A) $Molarity = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$
Since the volume of a solution is temperature-dependent,the molarity of the solution changes with a change in temperature.
Other concentration terms like molality,mass percentage,and mole fraction involve mass,which is independent of temperature.

(D) Given: Molarity $(M)$ = $0.8 \ M$,Density $(d)$ = $1.06 \ g \ cm^{-3}$,Molar mass of $H_2SO_4$ $(M_2)$ = $98 \ g \ mol^{-1}$.
Formula for molality $(m)$: $m = \frac{M \times 1000}{d \times 1000 - M \times M_2}$.
Substituting the values: $m = \frac{0.8 \times 1000}{1.06 \times 1000 - 0.8 \times 98}$.
$m = \frac{800}{1060 - 78.4} = \frac{800}{981.6} \approx 0.81499 \ m$.
Converting to $\times 10^{-3} \ m$: $0.81499 \times 10^3 \times 10^{-3} \ m \approx 815 \times 10^{-3} \ m$.

(B) The mole fraction of a component in a mixture is defined as the ratio of the number of moles of that component to the total number of moles of all components present in the solution.
The total number of moles in the solution is $n_{total} = n_A + n_B + n_C$.
The mole fraction of $C$ $(X_C)$ is given by:
$X_C = \frac{n_C}{n_A + n_B + n_C}$.

(B) The relationship between molality $(m)$ and molarity $(M)$ is given by the formula:
$m = \frac{1000 \times M}{(1000 \times d) - (M \times Mw_{\text{solute}})}$
Given:
$m = 3 \ molal$
$d = 1.12 \ g \ mL^{-1}$
$Mw_{\text{solute}} = 40 \ g \ mol^{-1}$
$M = x$
Substituting the values into the formula:
$3 = \frac{1000 \times x}{(1000 \times 1.12) - (x \times 40)}$
$3 = \frac{1000x}{1120 - 40x}$
$3(1120 - 40x) = 1000x$
$3360 - 120x = 1000x$
$3360 = 1120x$
$x = \frac{3360}{1120} = 3$
Therefore,the value of '$x$' is $3$.

(B) $3 \text{ M}$ solution means $3 \text{ moles}$ of $NaCl$ are present in $1 \text{ L}$ $(1000 \text{ mL})$ of solution.
Mass of solution = $\text{Volume} \times \text{Density} = 1000 \text{ mL} \times 1.25 \text{ g mL}^{-1} = 1250 \text{ g}$.
Mass of solute $(NaCl)$ = $3 \text{ mol} \times 58.5 \text{ g mol}^{-1} = 175.5 \text{ g}$.
Mass of solvent = $\text{Mass of solution} - \text{Mass of solute} = 1250 \text{ g} - 175.5 \text{ g} = 1074.5 \text{ g} = 1.0745 \text{ kg}$.
Molality $(m)$ = $\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{3}{1.0745} = 2.79 \text{ m}$.
Alternatively, using the formula:
$m = \frac{M \times 1000}{1000 \times d - M \times M_B}$
$m = \frac{3 \times 1000}{1000 \times 1.25 - 3 \times 58.5} = 2.79 \text{ m}$.

(C) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given $m = 4.44 \ m$,this means $4.44 \ mol$ of urea is dissolved in $1000 \ g$ $(1 \ kg)$ of water.
Moles of water $(n_{water})$ = $\frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Mole fraction of urea $(X_{urea})$ = $\frac{n_{urea}}{n_{urea} + n_{water}} = \frac{4.44}{4.44 + 55.56} = \frac{4.44}{60} = 0.074$.
We are given $X_{urea} = x \times 10^{-5}$.
$0.074 = 7400 \times 10^{-5}$.
Therefore,$x = 7400$.

(B) Mass of solute (ethyl alcohol) $= 1 \ mole \times 46 \ g \ mol^{-1} = 46 \ g$.
Mass of solvent (water) $= 9 \ mole \times 18 \ g \ mol^{-1} = 162 \ g$.
Total mass of solution $= 46 \ g + 162 \ g = 208 \ g$.
Mass percent of solute $= \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100$.
Mass percent $= \frac{46}{208} \times 100 = \frac{4600}{208} \approx 22.11 \%$.
The integer value is $22$.

(B) Given: Molarity $(M) = 0.2 \ M$,Volume $(V) = 500 \ mL = 0.5 \ L$,Density $(d) = 1.25 \ g/mL$,Molar mass of $CuSO_4 = 159.5 \ g/mol$.
Step $1$: Calculate moles of solute $(n)$:
$n = M \times V = 0.2 \times 0.5 = 0.1 \ mol$.
Step $2$: Calculate mass of solute $(x)$:
$x = n \times \text{Molar mass} = 0.1 \times 159.5 = 15.95 \ g$.
Step $3$: Calculate mass of solution:
$Mass_{soln} = V \times d = 500 \times 1.25 = 625 \ g$.
Step $4$: Calculate mass of solvent $(W_{solvent})$:
$W_{solvent} = Mass_{soln} - x = 625 - 15.95 = 609.05 \ g$.
Step $5$: Calculate molality $(m)$:
$m = \frac{n \times 1000}{W_{solvent} (in \ g)} = \frac{0.1 \times 1000}{609.05} \approx 0.16418 \ m$.
$m \approx 164 \times 10^{-3} \ m$.

(B) Let the molecular weight of the solute be $M_1$ and the solvent be $M_2$.
Given mole fraction of solute $X_1 = 0.1$,so mole fraction of solvent $X_2 = 0.9$.
For a solution,the relation between molarity $(M)$ and molality $(m)$ is given by $m = \frac{1000M}{1000d - MM_1}$,where $d$ is the density in $g \ cm^{-3}$.
Since $M = m$,we have $1 = \frac{1000}{1000d - MM_1}$.
$1000d - MM_1 = 1000$ $\Rightarrow 1000(2) - MM_1 = 1000$ $\Rightarrow MM_1 = 1000$.
Since $M = \frac{n_1}{V(L)}$,and $n_1 = \frac{mass}{M_1}$,we have $M = \frac{mass}{M_1 \times V(L)}$.
Given $d = \frac{mass}{V} = 2000 \ g \ L^{-1}$,so $mass = 2000 \times V$.
$M = \frac{2000 \times V \times w_1}{M_1 \times V} = \frac{2000 w_1}{M_1}$ where $w_1$ is the mass fraction of solute.
Using $X_1 = \frac{n_1}{n_1 + n_2} = 0.1$,we get $n_2 = 9n_1$.
$\frac{mass_2}{M_2} = 9 \times \frac{mass_1}{M_1} \Rightarrow \frac{mass_2}{mass_1} = 9 \times \frac{M_2}{M_1}$.
Since $M = m$,the mass of solvent is $1000 \ g$.
From $MM_1 = 1000$,we find $n_1 = 1 \ mol$.
Thus,$n_2 = 9 \ mol$.
Mass of solvent $= 9 \times M_2 = 1000 \ g \Rightarrow M_2 = \frac{1000}{9}$.
Mass of solute $= 1 \times M_1 = 1000 - (2000 - 1000) = 1000 \ g$ is incorrect; using $X_1 = 0.1$,$\frac{M_1}{M_2} = \frac{X_1}{X_2} \times \frac{n_2}{n_1} = \frac{0.1}{0.9} \times 9 = 1$ is wrong.
Correct approach: $\frac{M_1}{M_2} = \frac{X_1}{X_2} \times \frac{n_2}{n_1}$. With $M=m$,$d = 1 + \frac{M M_1}{1000} = 2$. $M M_1 = 1000$. $n_1 = M \times 1 = 1000/M_1$. $n_2 = (2000 - 1000)/M_2 = 1000/M_2$. $X_1 = \frac{1000/M_1}{1000/M_1 + 1000/M_2} = 0.1$ $\Rightarrow \frac{M_2}{M_1+M_2} = 0.1$ $\Rightarrow M_2 = 0.1M_1 + 0.1M_2$ $\Rightarrow 0.9M_2 = 0.1M_1$ $\Rightarrow \frac{M_1}{M_2} = 9$.

(C) Molarity is defined as the number of moles of solute dissolved in per litre of solution.
Molarity,$M = \frac{n}{V}$
$n =$ number of moles of solute = mass $/$ molar mass
$V =$ volume of solution
Mass of urea = $120 \ g$
Molar mass of urea = $60 \ g / mol$
Mass of water = $1000 \ g$
Density of solution,$\rho = 1.15 \ g / mL$
$n = \frac{120}{60} = 2 \ mol$
Total mass of solution = $m_{solute} + m_{solvent} = 120 \ g + 1000 \ g = 1120 \ g$
Volume of solution,$V = \frac{\text{mass}}{\text{density}} = \frac{1120 \ g}{1.15 \ g / mL} \approx 973.91 \ mL = 0.97391 \ L$
Molarity,$M = \frac{n}{V} = \frac{2 \ mol}{0.97391 \ L} \approx 2.05 \ M$
Hence,the correct option is $C$.

(D) Moles of water $(n_{H_2O})$ = $\frac{900}{18} = 50 \text{ mol}$.
Mole fraction of urea $(x_{urea})$ = $0.05$.
Using the formula $x_{urea} = \frac{n_{urea}}{n_{urea} + n_{H_2O}}$:
$0.05 = \frac{n_{urea}}{n_{urea} + 50}$.
$0.05n_{urea} + 2.5 = n_{urea} \implies 0.95n_{urea} = 2.5 \implies n_{urea} = \frac{2.5}{0.95} \approx 2.6316 \text{ mol}$.
Mass of urea = $2.6316 \text{ mol} \times 60 \text{ g mol}^{-1} = 157.896 \text{ g}$.
Total mass of solution = $157.896 \text{ g} + 900 \text{ g} = 1057.896 \text{ g}$.
Volume of solution $(V)$ = $\frac{\text{Mass}}{\text{Density}} = \frac{1057.896 \text{ g}}{1.2 \text{ g cm}^{-3}} = 881.58 \text{ mL} = 0.88158 \text{ L}$.
Molarity $(M)$ = $\frac{n_{urea}}{V(L)} = \frac{2.6316}{0.88158} \approx 2.98 \text{ M}$.

(C) Given: Molarity $(M)$ = $3.2 \ mol \ L^{-1}$.
Consider $1 \ L$ of the solution.
Number of moles of solute $(n)$ = $3.2 \ mol$.
Mass of solute = $n \times \text{Molar mass} = 3.2 \ mol \times 80 \ g \ mol^{-1} = 256 \ g$.
Mass of solution = $\text{Density of solution} \times \text{Volume}$. Since the density of the solvent is $0.4 \ g \ mL^{-1}$ and we assume no volume change,we consider the solvent volume as $1 \ L$ $(1000 \ mL)$.
Mass of solvent = $\text{Density} \times \text{Volume} = 0.4 \ g \ mL^{-1} \times 1000 \ mL = 400 \ g = 0.4 \ kg$.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{3.2 \ mol}{0.4 \ kg} = 8 \ m$.

(D) Given: Molarity $(M)$ = $3 \ M$,Density $(d)$ = $1.25 \ g/mL$,Molar mass of $NaCl$ $(M_w)$ = $58.5 \ g/mol$.
The formula for molality $(m)$ is: $m = \frac{M \times 1000}{1000 \times d - M \times M_w}$.
Substituting the values: $m = \frac{3 \times 1000}{1000 \times 1.25 - 3 \times 58.5}$.
$m = \frac{3000}{1250 - 175.5}$.
$m = \frac{3000}{1074.5} \approx 2.79 \ m$.

(C) Let the mass of the solution be $100 \ g$.
Since the solution is $70 \%$ (mass/mass),the mass of the solute $(X)$ is $70 \ g$.
The volume of the solution is calculated as: $V = \frac{\text{mass}}{\text{density}} = \frac{100 \ g}{1.25 \ g \ mL^{-1}} = 80 \ mL$.
The number of moles of the acid $(X)$ is: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{70 \ g}{70 \ g \ mol^{-1}} = 1 \ mol$.
Molarity $(M)$ is defined as: $M = \frac{n \times 1000}{V \text{ (in } mL)} = \frac{1 \times 1000}{80} = 12.5 \ M$.
Rounding to the nearest integer,we get $13 \ M$.

(A) The density of water is maximum at $4^{\circ}C$,which means the volume of water is minimum at $4^{\circ}C$.
As the temperature increases from $1^{\circ}C$ to $4^{\circ}C$,the volume of water decreases,causing the molarity $(M = n/V)$ to increase.
As the temperature increases from $4^{\circ}C$ to $25^{\circ}C$,the volume of water increases due to thermal expansion,causing the molarity to decrease.
Therefore,the molarity first increases and then decreases,which is represented by plot $A$.

(D) Solution $1$ consists of $10 \ mol$ of solute $x$ in $10 \ L$ of water,resulting in a concentration of $1 \ mol/L$.
Solution $2$ is prepared by taking $1 \ L$ of solution $1$ (which contains $1 \ mol$ of $x$ and $1 \ L$ of water) and adding $1 \ mol$ of solute $x$ and $1 \ L$ of water.
Total amount in solution $2$ is $2 \ mol$ of $x$ in $2 \ L$ of water,so the concentration remains $1 \ mol/L$.
Since the concentration and composition are identical,intensive properties like density and molar heat capacity remain unchanged.
Gibbs free energy $(G)$ is an extensive property,which depends on the total amount of matter in the system.
Since the total amount of substance in solution $1$ is greater than in solution $2$,the Gibbs free energy will change.

(A) Sea water is $6 \ M$ in $NaCl$,so $1000 \ mL$ of sea water contains $6 \ mol$ of $NaCl$.
$\text{Mass of solution} = \text{Volume} \times \text{density} = 1000 \times 2 = 2000 \ g$.
$\text{ppm} = \frac{\text{mass of } O_2}{\text{mass of solution}} \times 10^6$.
$5.8 = \frac{\text{mass of } O_2}{2000} \times 10^6$ $\Rightarrow \text{mass of } O_2 = 5.8 \times 2 \times 10^{-3} = 1.16 \times 10^{-2} \ g$.
$\text{Molality of } O_2 = \frac{\text{moles of } O_2}{\text{mass of solvent in } kg}$.
$\text{Mass of solvent} = \text{Mass of solution} - \text{Mass of } NaCl = 2000 - (6 \times 58.5) = 2000 - 351 = 1649 \ g = 1.649 \ kg$.
$\text{Moles of } O_2 = \frac{1.16 \times 10^{-2}}{32} = 0.03625 \times 10^{-2} = 3.625 \times 10^{-4} \ mol$.
$\text{Molality} = \frac{3.625 \times 10^{-4}}{1.649} \approx 2.197 \times 10^{-4} \ m$.
Thus,$x \approx 2$.

(A) . Humidity is a solution of water vapor (liquid) in air (gas),which is $Liquid \text{ in } gas$.
$B$. Alloys are homogeneous mixtures of metals,which are $Solid \text{ in } solid$.
$C$. Amalgams are solutions of mercury (liquid) in another metal (solid),which is $Liquid \text{ in } solid$.
$D$. Smoke consists of solid particles dispersed in air (gas),which is $Solid \text{ in } gas$.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(A) The volume of a solution changes with temperature.
Concentration terms that involve volume in their definition,such as Molarity $(M = \frac{\text{moles of solute}}{\text{volume of solution in L}})$,are dependent on temperature.
In contrast,Molality,Mole fraction,and Mass percentage are based on mass,which does not change with temperature.

(C) $0.25 \ m$ (molal) solution means $0.25 \ \text{moles}$ of urea are dissolved in $1000 \ g$ $(1 \ kg)$ of solvent.
The molar mass of urea $[NH_2CONH_2]$ is $14 + 2 + 12 + 16 + 14 + 2 = 60 \ g/mol$.
Mass of urea in $1000 \ g$ of solvent $= 0.25 \ \text{mol} \times 60 \ g/mol = 15 \ g$.
Total mass of the solution $= \text{Mass of solute} + \text{Mass of solvent} = 15 \ g + 1000 \ g = 1015 \ g = 1.015 \ kg$.
For $1.015 \ kg$ of solution,we need $15 \ g$ of urea.
Therefore,for $2.5 \ kg$ of solution,the mass of urea required $= (15 \ g / 1.015 \ kg) \times 2.5 \ kg \approx 36.94 \ g$.

(B) Given: Mass percentage of $C_2H_6O_2 = 28 \%$.
This means $28 \ g$ of $C_2H_6O_2$ is present in $100 \ g$ of the solution.
Mass of water = $100 \ g - 28 \ g = 72 \ g$.
Molar mass of $C_2H_6O_2 = (2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \ g/mol$.
Molar mass of $H_2O = 18 \ g/mol$.
Moles of $C_2H_6O_2 (n_1) = \frac{28}{62} \approx 0.4516 \ mol$.
Moles of $H_2O (n_2) = \frac{72}{18} = 4.000 \ mol$.
Mole fraction of $C_2H_6O_2 (x_1) = \frac{n_1}{n_1 + n_2} = \frac{0.4516}{0.4516 + 4.000} = \frac{0.4516}{4.4516} \approx 0.1014$.
Therefore,the mole fraction is approximately $0.101$.

(A) According to the dilution formula,$M_1 V_1 = M_2 V_2$,where $V_2$ is the final volume of the solution.
$V_2 = \frac{M_1 V_1}{M_2}$.
The volume of water to be added is $V_{\text{added}} = V_2 - V_1$.
Substituting $V_2$,we get $V_{\text{added}} = \frac{M_1 V_1}{M_2} - V_1$.
$V_{\text{added}} = V_1 (\frac{M_1}{M_2} - 1) = V_1 (\frac{M_1 - M_2}{M_2})$.

(B) $1$. The molar mass of acetic acid $(CH_3COOH)$ is calculated as: $(2 \times 12) + (4 \times 1) + (2 \times 16) = 60 \ g/mol$.
$2$. Given $4.0 \%$ $(w/w)$ means $4.0 \ g$ of acetic acid in $100 \ g$ of vinegar solution.
$3$. The volume of $100 \ g$ of solution is calculated using density: $V = \frac{mass}{density} = \frac{100 \ g}{1.02 \ g/mL} \approx 98.04 \ mL = 0.09804 \ L$.
$4$. Moles of acetic acid = $\frac{mass}{molar \ mass} = \frac{4.0 \ g}{60 \ g/mol} \approx 0.06667 \ mol$.
$5$. Molarity $(M) = \frac{moles \ of \ solute}{volume \ of \ solution \ in \ L} = \frac{0.06667 \ mol}{0.09804 \ L} \approx 0.68 \ M$.

(A) Step $1$: Calculate the moles of solute $(n_{solute})$.
$n_{solute} = \frac{\text{mass}}{\text{molar mass}} = \frac{394 \ g}{342 \ g \ mol^{-1}} \approx 1.152 \ mol$.
Step $2$: Calculate the moles of solvent $(n_{solvent})$.
$n_{solvent} = \frac{622 \ g}{18 \ g \ mol^{-1}} \approx 34.556 \ mol$.
Step $3$: Calculate the mole fraction of solute $(X_{solute})$.
$X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{1.152}{1.152 + 34.556} = \frac{1.152}{35.708} \approx 0.032$.

(D) An alloy is a homogeneous mixture of two or more metals,or a metal and a non-metal. Since both the solute and the solvent are in the solid state,an alloy is classified as a $Solid$ in $Solid$ solution.

(B) 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 $(mol/L)$.
Since volume is a component of the denominator,$Molarity$ changes with temperature.
In contrast,$Molality$,$Mole fraction$,and $Percent by mass$ are based on mass,which is independent of temperature.

(C) solution of ethyl alcohol in water is formed by dissolving a liquid solute (ethyl alcohol) in a liquid solvent (water). Therefore,it is classified as a $Liquid$ in $Liquid$ type of solution.

(A) In an amalgam of mercury with sodium,$Hg$ (mercury) is the solute and $Na$ (sodium) is the solvent.
Since mercury is a liquid and sodium is a solid,this forms a solution of a liquid in a solid.
Therefore,the correct classification is liquid in solid.

(D) true solution is a homogeneous mixture of two or more substances.
It is not a heterogeneous mixture.
In a true solution,the particle size of the solute is less than $10^{-9} \ m$ or $1 \ nm$.
Since the components are uniformly distributed at the molecular level,it exhibits consistent properties throughout.
Therefore,the statement in option $D$ is incorrect.

(B) Molar mass of $NaOH = 23 + 16 + 1 = 40 \ g \ mol^{-1}$.
Number of moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{2.0 \ g}{40 \ g \ mol^{-1}} = 0.05 \ mol$.
Volume of solution $= 500 \ cm^{3} = 0.5 \ L$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.05 \ mol}{0.5 \ L} = 0.1 \ mol \ dm^{-3}$.

(A) Given: Mass of solute $(CH_3COOH)$ $= 60 \ g$,Volume of solution $= 1 \ dm^3 = 1000 \ cm^3$,Density of solution $= 1.25 \ g / cm^3$.
Molar mass of $CH_3COOH = 12 + 3 + 12 + 16 + 16 + 1 = 60 \ g / mol$.
Moles of solute $= \frac{60 \ g}{60 \ g / mol} = 1 \ mol$.
Mass of solution $= \text{Volume} \times \text{Density} = 1000 \ cm^3 \times 1.25 \ g / cm^3 = 1250 \ g$.
Mass of solvent $= \text{Mass of solution} - \text{Mass of solute} = 1250 \ g - 60 \ g = 1190 \ g = 1.19 \ kg$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{1 \ mol}{1.19 \ kg} \approx 0.84 \ m$.
Note: Assuming the $1 \ dm^3$ refers to the volume of the solution,the calculated molality is approximately $0.84 \ m$. Given the options,$0.8 \ m$ is the closest approximation.

(A) Given: Mass of urea $= 15 \ g$,Molar mass of urea $= 60 \ g \ mol^{-1}$,Volume of solution $= 500 \ cm^3 = 0.5 \ L$.
Moles of urea $= \frac{\text{mass}}{\text{molar mass}} = \frac{15 \ g}{60 \ g \ mol^{-1}} = 0.25 \ mol$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.25 \ mol}{0.5 \ L} = 0.5 \ mol \ dm^{-3}$.

(D) Molarity $(M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in Litres}}$
Number of moles of $NaOH = \frac{\text{Mass}}{\text{Molar mass}} = \frac{5.0 \ g}{40 \ g \ mol^{-1}} = 0.125 \ mol$
Volume of solution $= 100 \ mL = 0.1 \ L$
Molarity $= \frac{0.125 \ mol}{0.1 \ L} = 1.25 \ mol \ L^{-1}$ (or $1.25 \ mol \ dm^{-3}$)
Therefore,the correct option is $D$.

(C) For $NaCl$: $n(Cl^-) = M \times V = 3.0 \ M \times 0.300 \ L = 0.9 \ mol$.
For $BaCl_{2}$: $BaCl_{2} \rightarrow Ba^{2+} + 2Cl^-$.
$n(Cl^-) = 2 \times M \times V = 2 \times 4.0 \ M \times 0.200 \ L = 1.6 \ mol$.
Total moles of $Cl^- = 0.9 \ mol + 1.6 \ mol = 2.5 \ mol$.
Total volume of solution = $300 \ mL + 200 \ mL = 500 \ mL = 0.5 \ L$.
Molar concentration of $Cl^- = \frac{\text{Total moles}}{\text{Total volume}} = \frac{2.5 \ mol}{0.5 \ L} = 5.0 \ M$.

(B) Given: Volume of solution $= 1500 \ cm^{3}$,Density of solution $= 1.052 \ g/cm^{3}$,Mass of urea (solute) $= 18 \ g$,Molar mass of urea $= 60 \ g/mol$.
Mass of solution $= \text{Volume} \times \text{Density} = 1500 \ cm^{3} \times 1.052 \ g/cm^{3} = 1578 \ g$.
Mass of solvent (water) $= \text{Mass of solution} - \text{Mass of solute} = 1578 \ g - 18 \ g = 1560 \ g = 1.560 \ kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{18 / 60}{1.560} = \frac{0.3}{1.560} \approx 0.192 \ m$.

(B) Molality is defined as the number of moles of solute per kilogram of solvent.
Formula: $\text{Molality} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}}$
Given: Mass of glucose $= 450 \ mg = 0.45 \ g$,Molar mass of glucose $(C_6H_{12}O_6) = 180 \ g/mol$,Mass of solvent $= 100 \ g = 0.1 \ kg$.
Calculation: $\text{Molality} = \frac{0.45 \ g}{180 \ g/mol \times 0.1 \ kg} = \frac{0.45}{18} = 0.025 \ m$.

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

(B) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Number of moles of solute $(n_2)$ = $\frac{W_2}{M_2}$.
Mass of solvent in kg = $\frac{W_1}{1000}$.
Therefore,$m = \frac{n_2}{\text{mass of solvent in kg}} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000 \times W_2}{M_2 \times W_1}$.

(B) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
$M = \frac{n_{solute}}{V_{solution(L)}}$
Since the volume of a solution $(V)$ is dependent on temperature (due to thermal expansion or contraction),a change in temperature will change the volume of the solution.
Therefore,molarity is affected by a change in temperature.

(A) Mass of sugar $= 3.42 \times 10^{-2} \ kg = 34.2 \ g$.
Mass of sugar syrup $= 234.2 \ g$.
Mass of solvent (water) $= 234.2 \ g - 34.2 \ g = 200 \ g = 0.2 \ kg$.
Moles of sugar $= \frac{\text{Mass}}{\text{Molar mass}} = \frac{34.2 \ g}{342 \ g \ mol^{-1}} = 0.1 \ mol$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$.

(B) $2$ molal aqueous solution means $2$ moles of solute are present in $1 \ kg$ $(1000 \ g)$ of water.
Moles of solute $(n_{solute})$ $= 2 \ mol$.
Moles of solvent (water,$n_{solvent}$) $= \frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Total moles $= n_{solute} + n_{solvent} = 2 + 55.56 = 57.56 \ mol$.
Mole fraction of solute $(x_{solute})$ $= \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{2}{57.56} \approx 0.0347$.

(A) Mass percentage of solute $= 0.25 \%$. This means $0.25 \ g$ of solute is present in $100 \ g$ of solution.
Mass of solvent $= \text{Mass of solution} - \text{Mass of solute} = 100 \ g - 0.25 \ g = 99.75 \ g$.
Thus,$0.25 \ g$ of solute is present in $99.75 \ g$ of solvent.
For $1.25 \ g$ of solute,the mass of solvent required is $\frac{99.75 \ g \text{ solvent}}{0.25 \ g \text{ solute}} \times 1.25 \ g \text{ solute} = 498.75 \ g$.

(D) Step $1$: Calculate the number of moles of urea. Given mass $= 300 \ mg = 0.3 \ g$. Molar mass $= 60 \ g/mol$. Moles $(n) = \frac{0.3 \ g}{60 \ g/mol} = 0.005 \ mol$.
Step $2$: Calculate the mass of the solvent in $kg$. Mass of water $= 30 \ g = 0.03 \ kg$.
Step $3$: Calculate molality $(m)$. Molality $= \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{0.005 \ mol}{0.03 \ kg} = 0.166 \ m$.

(B) Mass of $NaOH$ $(m) = 3.2 \ g$
Volume of solution $(V) = 250 \ cm^{3} = 0.25 \ L$
Molar mass of $NaOH$ $(M) = 40 \ g \ mol^{-1}$
Number of moles $(n) = \frac{m}{M} = \frac{3.2 \ g}{40 \ g \ mol^{-1}} = 0.08 \ mol$
Molarity $(M) = \frac{n}{V(L)} = \frac{0.08 \ mol}{0.25 \ L} = 0.32 \ mol \ dm^{-3}$

(D) Step $1$: Calculate the number of moles of $NaOH$ $(n)$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.8 \ g}{40 \ g \ mol^{-1}} = 0.02 \ mol$
Step $2$: Convert the volume of the solution from $cm^{3}$ to $L$:
$V = 150 \ cm^{3} = 0.15 \ L$
Step $3$: Calculate the molarity $(M)$:
$M = \frac{n}{V(L)} = \frac{0.02 \ mol}{0.15 \ L} = 0.1333 \ mol \ dm^{-3}$

(A) Given: Mass of sugar $= 34.2 \ g$,Molar mass of sugar $= 342 \ g/mol$.
Number of moles of sugar $(n_{sugar}) = \frac{34.2}{342} = 0.1 \ mol$.
Mass of water $= 1.8 \times 10^{2} \ g = 180 \ g$,Molar mass of water $= 18 \ g/mol$.
Number of moles of water $(n_{water}) = \frac{180}{18} = 10 \ mol$.
Mole fraction of sugar $(x_{sugar}) = \frac{n_{sugar}}{n_{sugar} + n_{water}} = \frac{0.1}{0.1 + 10} = \frac{0.1}{10.1} \approx 0.0099$.

(A) Molarity $(M)$ is defined as the number of moles of solute present in $1 \ dm^3$ (or $1 \ L$) of solution.
$Molarity (M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in } dm^3}$