Solutions Test -23

Given that:

Mass of \({CO}({NO})_{3} .6 {H}_{2} {O}=30~g\)

Volume of solution \(=4.3 {~L}\)

We know that,

Molarity \(=\frac{\text { Moles of Solute }}{\text { Volume of solution in litre }}\)

Molar mass of \({CO}({NO})_{3} \cdot 6 {H}_{2} {O}=59+2(14+3 \times 16)+6 \times 18=291 {~g}\) \({mol}^{-1}\)

Therefore, 

Moles of Moles of \({CO}({NO})_{3} \cdot 6 {H}_{2} {O}=\frac{\text{Mass}}{\text{Molar Mass}}\)

\(=\frac{30}{291} {~mol}\)

\(=0.103 {~mol}\)

Therefore, 

Molarity \(=\frac{0.103 {~mol}}{4.3 L}\)

\(=0.023 {M}\)

Hence, the correct option is (C).

Given that:

Mass of acetic acid \(\left(w_{1}\right)=75 {~g}\)

Lowering of the melting point, \(\Delta T_{f}=(1.5+273)-(0+273)=1.5 {~K}\)

Molar mass of ascorbic acid \(\left({C}_{6} {H}_{8} {O}_{6}\right)\) will be, 

\({M}_{2}=6 \times 12+8 \times 1+6 \times 16\)

\(=176 {~g}\)\({mol}^{-1}\)

We know that:

\(\Delta T_{f}=\frac{K_{f} \times w_{2} \times 1000}{M_{2} \times w_{1}}\)

Or, \( w_{2}=\frac{\Delta T_{f} \times M_{2} \times w_{1}}{K_{f} \times 1000}\)

\(=\frac{1.5 \times 176 \times 75}{3.9 \times 1000}\)

\(=5.08 {~g}\)

Hence, the correct option is (D).

Potassium dichromate is a primary standard.

• The primary standard is a compound of sufficient purity from which standard solutions of known normalities can be prepared by direct weighing of it and diluting to a defined volume of solution.
• Potassium Dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right)\) is suitable to be used as a primary standard. It cannot be obtained in very pure form. It readily reacts with any traces of organic material or any other reducing substance in water.

Hence, the correct option is (D).

Given that:

Mass of glucose \(=36~g\)

Molar mass of Glucose \((C_6H_{12}O_6 )=12 \times 6 + 1 \times 12 + 16 \times 6=180\)

\(\pi_{1}=4.98\) bar

\(\pi_{2}=1.52\) bar

Mole fraction of glucose, \(C_{1}=\frac{\text{Mass}}{\text{Molar Mass}}\)

\(=\frac{36}{180}\)

\(C_{2}=? \)

Now, according to van't hoff equation:

\(\pi=\mathrm{CRT}\)

Putting the values in above equation, we get:

\(4.98=\left(\frac{36}{180}\right) \mathrm{RT}\quad \cdots\) (1)

\(1.52=\mathrm{C}_{2} \mathrm{RT}\quad \cdots\) (2)

Now dividing equation (2) by (1), we get:

\(\frac{\left(C_{2} \times 180\right)} {36}=\frac{1.52}{4.98}\)

or, \(C_{2}=0.061\)

Therefore concentration of second solution is \(0.061 \mathrm{M}\).

Hence, the correct option is (A).

An aqueous solution of hydrochloric acid shows negative deviation from Raoult's law.

Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present, i.e.,

\(\left[\mathrm{P}_{\text {solution }}=\mathrm{X}_{\text {solvent }} \mathrm{P}_{\text {solvent }}^{\circ}\right]\)

If the attraction between different molecules, for example between \(\mathrm{HCl}\) and \(\mathrm{H}_{2} \mathrm{O}\) molecules, is stronger, the escaping tendency from the solution to the vapour phase will be smaller, then the partial vapour pressure will be smaller than predicted by Raoult's law and the system exhibits a negative deviation.

Hence, the correct option is (B).

Given that:

The solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass.

Therefore,

Total amount of solute present in the mixture will be given by,

\(300 \times \frac{25}{100}+400 \times \frac{40}{100}\)

\(=75+160\)

\(=235 \mathrm{~g}\)

Total amount of solution \(=300+400=700 \mathrm{~g}\)

Therefore, 

Mass percentage of the solute in the resulting solution \(=\frac{\text{Total amount of solute}}{\text{Total amount of solution}} \times100\)

\(=\frac{235}{700} \times100\)

\(=33.57 \%\)

Then, mass percentage of the solvent in the resulting solution will be:

\(=(100-33.57) \%\)

\(=66.43 \%\)

Hence, the correct option is (C).

Given that:

Mass of liquid \(A=100~g\) 

Molar mass of liquid \(A=140~g~mol^{-1}\) 

Mass of liquid \(B=1000~g\) 

Molar mass of liquid \(B=180~g~mol^{-1}\) 

Number of moles of liquid \({A}\),

\(n_{A}=\frac{100}{140}\)

\(=0.714 {~mol}\)

Number of moles of liquid \({B}\), 

\(n_{B}=\frac{1000}{180}\)

\(=5.556 {~mol}\)

Then, mole fraction of \({A}\),

\(x_{A}=\frac{n_{A}}{n_{A}+n_{B}}\)

\(=\frac{0.714}{0.714+5.556}\)

\(=\frac{0.714}{0.714+5.556}\)

\(=0.114\)

And, mole fraction of \({B}\),

\({x}_{{B}}=1-0.114\)

\(=0.886\)

Vapour pressure of pure liquid \({B}\),

\(p_{B}^{\circ}=500\) torr

Therefore, vapour pressure of liquid \(B\) in the solution,

\(p_{B}=p_{B}^{\circ} x_{B}\)

\(=500 \times 0.886\)

\(=443\) torr

Total vapour pressure of the solution, 

\(p_{\text {total }}=475\) torr

Therefore, Vapour pressure of liquid A in the solution,

\(p_{A}=p_{\text {total }}-p_{B}\)

\(=475-443\)

\(=32\) torr

Now, 

\(p_{A}=p_{A}^{\circ} x_{A}\)

Or, \( p_{A}^{\circ}=\frac{p_{A}}{x_{A}}\)

\(=\frac{32}{0.114}\)

\(=280.7\) torr

Hence, the correct option is (D).

Molar mass of \({KI}=39+127=166 {~g} {~mol}^{-1}\)

\(20 \%\) aqueous solution of KI means \(20 {~g}\) of \({KI}\) is present in \(100 {~g}\) of solution.

Therefore, 

Mass of \(KI=20 {~g}\)

That is, 

\(20 {~g}\) of \({KI}\) is present in \((100-20) {g}\) of water \(=80 {~g}\) of water

We know that:

Molality of the solution \(=\frac{\text { Moles of } K I}{\text { Mass of water in } {kg}}\)

\(=\frac{\frac{\text {Mass of KI}}{\text {Molar Mass of KI}} }{\text { Mass of water in } {kg}}\)

\(=\frac{\frac{20}{166}}{0.08} {~m}\)

\(=1.506 {~m}\)

\(=1.51 {~m}\)

Hence, the correct option is (A).

Given:

Weight of Urea \(=12 {~g}\)

And, weight of sucrose \(=68.4 {~g}\)

We know that:

Number of moles \(=\frac{\text{Weight}}{\text{Molar mass}}\)

Therefore,

Moles of urea \(=\frac{12}{60}=0.2\)

(Molar mass of Urea is \(60 {gm}\). )

Moles of sucrose \(=\frac{68.4}{342}=0.2\)

(Molar mass of sucrose is \(342 {gm}\). )

Since, there are equal moles of solute in equal volumes of water, the mole fraction is the same. 

As per Raoult's law, the relative lowering of pressure will also be the same.

Hence, the correct option is (A).

In an endothermic process, solubility increases with increase in temperature. 

In saturated state, there exists the following equilibrium:

solute + solvent \(\rightleftharpoons\) solution

In endothermic reaction, \(\Delta \mathrm{H}=\) positive. When temperature is increased, more heat is added to reactant side. According to Le-Chatliers principle, the equilibrium then shifts to right. Therefore, solubility increases with increase in temperature.

Hence, the correct option is (A).

During osmosis, the flow of water through a semi-permeable membrane is from a solution having lower concentration only.

Osmosis is a process by which molecules of a solvent tend to pass through a semipermeable membrane from a less concentrated solution into a more concentrated one. The flow of water through a semipermeable membrane is from both sides of semipermeable membrane with unequal flow rates.

Hence, the correct option is (D).

A solution of acetone in ethanol shows a positive deviation from Raoult's law.

• It is due to miscibility of these two liquids with a difference of polarity and length of the hydrocarbon chain. 
• Positive derivation occurs when vapour pressure of the component is greater than expected value.
  
• Acetone and ethanol both the components escape easily showing higher vapour pressure than the expected value.

Hence, the correct option is (B).

A student slowly mixes salt into 25 ml of water until no more salt dissolves in it. The student could make more salt dissolve in the solution by heating it.

• As no more salt is dissolving in the solution, the solution has became saturated solution. 
• The student can dissolve more solution by heating the solution as on heating the saturated solution become supersaturated. 
• After, the solution become supersaturated, no more salt can be dissolved, it will precipitate out.

Given:

Mass of acetic acid, \(\mathrm{w_1}=75 \mathrm{~g}\) 

Molar mass of ascorbic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\),

\(\mathrm{M_2}=6 \times 12+8 \times 1+6 \times 16\)

\(=176 \mathrm{~g} \mathrm{~mol}^{-1}\)

Lowering of melting point, 

\(\Delta \mathrm{T}_{\mathrm{f}}=(1.5+273)-(0+273)=1.5 \mathrm{~K}\)

We know that:

\(\Delta \mathrm{T}_{\mathrm{f}}=\frac{\mathrm{K}_{\mathrm{f}} \times 1000 \times \mathrm{w}_{2}}{\mathrm{M}_{2} \times \mathrm{w}_{1}}\)

\(\mathrm{w}_{2}=\frac{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{M}_{2} \times \mathrm{w}_{1}}{\mathrm{~K}_{\mathrm{f}} \times 1000}\)

\(\mathrm{w}_{2}=\frac{1.5 \times 176 \times 75}{3.9 \times 1000}\)

\(\mathrm{w}_{2}=5.08 \mathrm{~g}\)

Therefore, \(5.08 \mathrm{~g}\) of ascorbic acid is needed to be dissolved.

Hence, the correct option is (C).

Depression in freezing point is a colligative property which depends upon the amount of the solute.

\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}\) 

Thus, for a given solvent and given concentration, \(\Delta T_{f}\) is directly porportional to \(i\) (Van't Hoff factor) i.e. maximum \(T_{f}\) (and hence lowest freezing point) will correspond to maximum value of \(i \).

(A) \(A l_{2}\left(S O_{4}\right)_{3} \stackrel{H_{2} O}{\longrightarrow} 2 A l^{3+}+3 S O_{4}^{2-}\)

Here, Van't Hoff factor, i = 5 

(B) \(C_{5} H_{10} O_{8} \stackrel{H_{2} O}{\longrightarrow}\) No ionization 

Here, Van't Hoff factor, i = 1 

(C) \(K I \stackrel{H_{2} O}{\longrightarrow} K^{+}+I^{-}\)

Here, Van't Hoff factor, i = 2 

(D) \(C_{12} H_{22} O_{11} \stackrel{H_{2} O}{\longrightarrow}\) No ionization 

Here, Van't Hoff factor, i = 2 

Therefore, \(0.10 M(\approx 0.10 \mathrm{~m})\) aqueous solution will have the lowest feezing point.

Hence, the correct option is (A).

Given:

Weight of solute \(({w})=1 {~g}\)

Weight of solvent \(({W})=75 {~g}\)

Boiling point of solution \(=100.114^{\circ} {C}\)

Boiling point of solvent \(=100^{\circ} {C}\)

Therefore, 

\(\triangle {T}=100.114-100=0.114^{\circ} {C}\)

Molecular weight of solute \(({m})=60.1\)

Boiling point elevation constant \(({K})=?\)

We know that:

\({m}=\frac{1000 \times {K} \times {w}}{\Delta {T} \times {W}}\)

or, \({K}=\frac{{m} \times \Delta {T} \times {W}}{1000 \times {w}}\)

\(=\frac{60.1 \times 0.114 \times 75}{1000 \times 1}\)

\(=\frac{513.8}{1000}\)

\(K=0.513\)

Hence, the correct option is (B).

Given:

Mole fraction of solute in first solution \(= 0.2\)

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of solute, i.e.,

\(\frac{p^{\circ}-p}{p^{\circ}}=\frac{n}{n+N}\)

or, \(\frac{\Delta p}{p^{\circ}}=\frac{n}{n+N} \)

The decrease in vapour pressure is \(10\) mm Hg:

\(\frac{10}{p^{\circ}}=0.2\)

\(\therefore p^{\circ}=50 \mathrm{~mm}\)  ....(1)

For other solution of same solvent, the decrease in vapor pressure is \(20\) mm Hg:

\(\frac{20}{p^{\circ}}=\frac{n}{n+N}\) 

or, \(\frac{20}{50}=\frac{n}{n+N}\)   (from (1))

\(0.4=\frac{n}{n+N}\) (mole fraction of solute)

\(\because\) Mole fraction of solvent \(+\) mole fraction of solute \(=1\)

So, mole fraction of solvent \(=1-0.4=0.6\)

Hence, the correct option is (B).

Given that:

Weight of solute \(=1.00~g\)

Molar Mass of the solute \(=250~g~mol^{-1}\)

Weight of solvent \(=51.2~g=0.0512~kg\)

We know that:

Molality of non- electrolyte solute \( =\frac{\frac{\text { weight of solute in gram }}{\text { molecular weight of solute }}}{\text { weight of solvent in } {kg}}\)

\(=\frac{\frac{1}{250}}{0.0512}\)

\(=\frac{1}{250 \times 0.0512}\)

\(=0.0781 {~m}\)

As we know:

\(\Delta {T}_{{f}}={K}_{{f}} \times\) molality of solution

\(=5.12 \times 0.0781\)

\(=0.4 {~K}\)

Hence, the correct option is (A).

Assume the mass of benzene be \(30 {~g}\) in the total mass of the solution of \(100 {~g}\). 

Mass of \({CCl}_{4}=(100-30) {g}\)

\(=70 {~g}\)

Molar mass of benzene \(\left({C}_{6} {H}_{6}\right)=(6 \times 12+6 \times 1) {g} {mol}^{-1}\)

\(=78 {~g} {~mol}^{-1}\)

Therefore, 

Number of moles of \({C}_{6} {H}_{6} = \frac{\text{Mass}}{\text{Molar Mass}}\)

\(=\frac{30}{78} {~mol}\)

\(=0.3846 {~mol}\)

Molar mass of \(({CCl}_{4})=(1 \times 12+4 \times 355){~g} {~mol}^{-1}\)

\(=154 {~g} {~mol}^{-1}\)

Therefore, 

Number of moles of \({CCl}_{4} = \frac{\text{Mass}}{\text{Molar Mass}}\)

\( =\frac{70}{154} ~mol\) 

\(=0.4545 {~mol}\)

Therefore, 

Mole fraction of benzene \(=\frac{\text { Number of moles of } C_{6} H_{6}}{\text { Number of moles of } C_{6} H_{6}+\text { Number of moles of } {CCl}_{4}}\)

\(=\frac{0.3846}{0.3846+0.4545}\)

\(=0.458\)

Hence, the correct option is (D).

Given that:

Mass of water, \(w_{1}=500 {~g}\)

Elevation constant, \(K_b=0.52~K~kg~mol^{-1}\)

Boiling point of sucrose = 100°C

Boiling point of water = 99.63°C

Therefore, elevation of boiling point, \(\Delta T_{b}=(100+273)-(99.63+273)\)

= 0.37 K

Molar mass of sucrose \(\left({C}_{12} {H}_{22} {O}_{11}\right)\) will be,  

\(M_{2}=12 \times 12+22 \times 1+11 \times 16\)

\(=342\)

We know that:

\(\Delta T_{b}=\frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}\)

Or, \(w_{2}=\frac{\Delta T_{b} \times M_{2} \times w_{1}}{K_{b} \times 1000}\)

\(=\frac{0.37 \times 342 \times 500}{0.52 \times 1000}\)

\(=121.67 {~g}\)

Hence, the correct option is (D).

Given that:

Vapour pressure of heptane, \(p_{1}^{\circ}=105.2 {kPa}\)

Vapour pressure of octane, \(p_{2}^{\circ}=46.8 {kPa}\)

We know that,

Number of moles \(=\frac{\text{Mass}}{\text{Molar Mass}}\) 

Molar mass of heptane \(\left({C}_{7} {H}_{16}\right)=7 \times 12+16 \times 1=100 {~g} {~mol}^{-1}\)

Therefore, number of moles of heptane \(=\frac{26}{100}=0.26 {~mol}\)

Molar mass of octane \(\left({C}_{8} {H}_{18}\right)=8 \times 12+18 \times 1=114 {~g} {~mol}^{-1}\)

Therefore, number of moles of octane \(=\frac{35}{114}=0.31 {~mol}\)

Mole fraction of heptane, \(x_{1}=\frac{\text{Number of moles of heptane}}{\text{Number of moles of heptane+Number of moles of octane}}\)

\(=\frac{0.26}{0.26+0.31}\)

\(=0.456\)

Therefore, mole fraction of octane, \(x_{2}=1-0.456=0.544\)

Now, partial pressure of heptane, 

\(p_{1}=x_{1} p_{1}^{\circ}\)

\(=0.456 \times 105.2\)

\(=47.97 {kPa}\)

Partial pressure of octane, 

\(p_{2}=x_{2} p_{2}^{\circ}\)

\(=0.544 \times 46.8\)

\(=25.46 {kPa}\)

Therefore, vapour pressure of solution, 

\(p_{\text {total }}=p_{1}+p_{2}\)

\(=47.97+25.46\)

\(=73.43 {kPa}\)

Hence, the correct option is (B).

Benzene and toulene will form an ideal solution.

• The substances that have similar structures and polarities form nearly ideal solution. 
• Since both benzene and toluene are non-polar, operating intermolecular forces are almost similar. Therefore, they form an ideal solution. 
• An ideal solution is a homogeneous mixture of substances that has physical properties linearly related to its pure components or obeys Raoult’s law.  

Hence, the correct option is (B).

Given:

Vapour pressure of liquid A,

\(P_{A} =120~ mm\)

Vapour pressure of liquid B,

\(P_{B} =180~ mm\)

Number of moles of liquid A, 

\({n}_{{A}}=2\)

Number of moles of liquid B, 

\({n}_{{B}}=3\)

Therefore,

Mole fraction of liquid A, 

\({X}_{{A}}=\frac{n_A}{n_A+n_B}\)

\(=\frac{2}{5}\)

Mole fraction of liquid B, 

\({X}_{{B}}=\frac{n_B}{n_A+n_B}\)

\(= \frac{3}{5}\)

We know that:

Vapor Pressure of solution, 

\({P}={X}_{{A}} {P}_{{A}}+{X}_{{B}} {P}_{{B}}\)

\(=\frac{2}{5} \times 120+\frac{3}{5} \times 180\)

\(=48+108\)

\(P=156 {~mm} {~Hg}\)

Hence, the correct option is (A).

It is given that:

Volume of water \(({V})=450 {~mL}=0.45 {~L}\)

Temperature \((T)=37+273=310 {~K}\)

Number of moles of the polymer, \(n=\frac{\text{Mass of polymer}}{\text{Molar Mass of polymer}}\)

\(=\frac{1}{185000}\) mol

We know that:

Osmotic pressure, \(\pi=\frac{n}{V} R T\)

where, \(R=8.314 \times 10^{3} {PaL} {K}^{-1} {~mol}^{-1}\)

\(=\frac{1}{185000} {~mol} \times \frac{1}{0.45 L} \times 8.314 \times 10^{3} {PaL} {K}^{-1} {~mol}^{-1} \times 310 {~K}\)

\(=30.98 {~Pa}\)

\(=31 {~Pa}\)

Hence, the correct option is (A).

It is given that:

The solution contains \(10 \mathrm{~g}\) per \(d m^{3}\) urea.

And, we know that:

The chemical formula of urea is \(\mathrm{CH}_{4} \mathrm{~N}_{2} \mathrm{O}\).

So, molecular mass of urea \(=12+4 \times(1)+2 \times(14)+16\)

\(=60 \mathrm{~gmol}^{-1}\)

So, the molarity of urea = \(\frac{\text{mass concentration}}{\text{molar mass}}\) 

\(=\left(\frac{10 g / d m^{3}}{60}\right)\)

\(=\frac{1}{6} m o l / d m^{3}\) 

Let's assume the molar mass of the non-volatile solution is '\(\mathrm{m}\)'.

So, the molarity of non- volatile solute =\(\frac{\text{mass concentration}}{\text{molar mass}}\)

\(=\frac{50 g / d m^{3}}{m}\)

Both solutions are isotonic with each other means concentration of both solutions are same:

Molarity of urea = Molarity of non-volatile solution

\(\frac{1}{6} \mathrm{~mol} / \mathrm{dm}^{3}=\frac{50 \mathrm{~g} / \mathrm{dm}^{3}}{\mathrm{~m}}\)

By solving the above equation we get the value of '\(m\)' as:

\(m=50 \times 6 \mathrm{~gmol}^{-1}\)

\(=300 \mathrm{~gmol}^{-1}\)

So, the molecular mass of non-volatile solute is \(300 \mathrm{~gmol}^{-1}\). 

Hence, the correct option is (A).

Let compound be B.

Water be A. 

\(\Delta {T}_{{f}}={T}_{{f}}^{0}-{T}_{{FS}}\)

Here, \({T}_{{f}}^{0} =0\)

Therefore,

\(\Delta {T}_{{f}}=0^{\circ}-(-0.465)\)

\(\Delta {T}_{{f}}=0.465^{\circ} {C}\)

Given:

\({K}_{{f}}=1.86~ {kg} ~{K} {mol^{-1}}\)

Weight of compound, \(w_{{B}}=1.8 {gm}\)

Let Molecular weight of compound, \( {M}_{{B}}={x}\)

Weight of water, \(w_{{A}}=40 {~g}\)

We know that:

\( \Delta {T}_{{f}}={K}_{{f}} \cdot \frac{w_{{B}}}{{M}_{{B}}} \times \frac{1000}{w_{{A}}}\)

\(0.465=\frac{1.86 \times 1.8 \times 1000}{x \times 40}\)

\({x}=180 {~g} / {mol}\)

So, empirical mass of \((CH_2O)\) \(=(12 \times 1)+(2 \times 1)+(16 \times 1)\)

\(=30\)

\({n}=\frac{\text{Molecular Mass}}{\text{Empirical formula mass}} \)

\( =\frac{180}{30}\)

\(n=6\)

So, molecular formula \(=(\)Emperical formula\()_n\)

\(= \left({CH}_{2} {O}\right)_6\)

\(={C}_{6} {H}_{12} {O}_{6}\)

Hence, the correct option is (D).

Given that:

Vapour pressure of the solution at normal boiling point, \(p_{1}=1.004\) bar

Vapour pressure of pure water at normal boiling point, \(p_{1}^{\circ}=1.013\) bar

Mass of solute, \(w_{2}=2 {~g}\)

Mass of solvent (water), \( w_{1} =100-2=98 \mathrm{~g}\)

Molar Mass of solvent (water), 

\({M}_{1}(H_2O)=2\times1+16\times1=18 {~g} {~mol}^{-1}\)

According to Raoult's law,

\(\frac{p_{1}^{\circ}-p_{1}}{p_{1}^{\circ}}=\frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}\)

\(\Rightarrow \frac{1.013-1.004}{1.013}=\frac{2 \times 18}{M_{2} \times 98}\)

\(\Rightarrow \frac{0.009}{1.013}=\frac{2 \times 18}{M_{2} \times 98}\)

\(\Rightarrow M_{2}=\frac{1.013 \times 2 \times 18}{0.009 \times 98}\)

\(\Rightarrow M_{2}=41.35 {~g} {~mol}^{-1}\)

Hence, the correct option is (C).

We know that:

Freezing point \(\propto \frac{1}{\text { Van't Hoff factor }}\)

(A) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl} \rightarrow C_{6} \mathrm{H}_{5} N H_{3}^{\oplus}+C l^{\ominus}\)

Here, Van't Hoff factor, i = 2 

(B)\(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{\ominus}\)

Here, Van't Hoff factor, i = 3 

(C)\(\mathrm{} \mathrm{La}\left(\mathrm{NO}_{3}\right)_{3} \rightarrow \mathrm{La}^{3+}+3 \mathrm{NO}_{3}^{\ominus}\)

Here, Van't Hoff factor, i = 4 

(D) Glucose does not dissociate. So, it has the minimum number of particles, and therefore, it shows minimum depression in freezing point. So, it has the maximum freezing point.

Hence, the correct option is (D).

Given that:

Mass of benzene, \((C_6 H_6)=22\) 

Mass of Carbon tetra chloride, \((CCl_4)=122\) 

We know that:

Mass percentage of Benzene \(\left({C}_{6} {H}_{6}\right)=\frac{\text { Mass of } C_{6} H_{6}}{\text { Total mass of the solution }} \times 100\)

\(=\frac{\text { Mass of } C_{6} H_{6}}{\text { Mass of } C_{6} H_{6}+\text { Mass of } C C l_{4}} \times 100\)

\(=\frac{22}{22+122} \times 100\)

\(=15.28 \%\)

Mass percentage of Carbon Tetrachloride \(\left({CCl}_{4}\right) =\frac{\text { Mass of } C C l_{4}}{\text { Total mass of the solution }} \times 100\)

\(=\frac{\text { Mass of } {CCl}_{4}}{\text { Mass of } C_{6} H_{6}+\text { Mass of } {CCl}_{4}} \times 100\)

\(=\frac{122}{22+122} \times 100\)

\(=84.72 \%\)

Hence, the correct option is (B).

0.15 M solution of benzoic acid in methanol means,

1000 mL of solution contains 0.15 mol of benzoic acid.

Therefore, \(250 {~mL}\) of solution contains \(\frac{0.15 \times 250}{1000}\) mol of benzoic acid

\(=0.0375\) mol of benzoic acid

Molar mass of benzoic acid \(\left({C}_{6} {H}_{5} {COOH}\right)=7 \times 12+6 \times 1+2 \times 16=122 {~g} {~mol}^{-1}\)

Therefore, required benzoic acid \(=0.0375 {~mol} \times 122 {~g} {~mol}^{-1}=4.575 {~g}\)

Hence, the correct option is (D).