Solutions practice · Hard

For an ideal solution of ${A}$ and ${B}, Y_{{A}}$ is the mole fraction of ${A}$ in the vapour phase at equilibrium. Which of the following plot should be linear?

Cyclohexane and ethanol at a particular temperature have vapour pressure of $280 {~mm}$ and $168 {~mm}$ respectively. If these two solutions having mole fraction value of cyclohexane equal to 0.32 are mixed and the mixture has a total vapour pressure of $376 {~mm}$, will the mixture be an ideal solution?

A solution of $1.25 \mathrm{~g}$ of $P$ in $50 \mathrm{~g}$ of water lowers the freezing point of water by $0.3^{\circ} \mathrm{C}$. Molar mass of $P$ is 94. $K_{f(\text { water })}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The degree of association of $P$ in water is

The total vapour pressure of a 4 mole $\%$ solution of $\mathrm{NH}_3$ in water at $293 \mathrm{~K}$ is 50.0 torr. The vapour pressure of pure water is 17.0 torr at this temperature. Applying Henry's and Raoult's laws, the total vapour pressure for a 5 mole $\%$ solution is

Liquid $M$ and liquid $N$ form an ideal solution. The vapour pressures of pure liquids $M$ and $N$ are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is
${x_M} =$ mole fraction of $M$ in solution; $\,\,\,\,\,\,\,\,\,\,\,{x_N} =$ mole fraction of $N$ in solution;
${y_M} =$ mole fraction of $M$ in vapour phase; $\,\,\,\,\,\,\,\,\,\,\,\,\,\,{y_N} =$ mole fraction of $N$ in vapour phase;

$20\;g$ of binary electrolyte (mol. wt. =100) are dissolved in $500\;g$ of water. The depression in freezing point of the solution is $0.74^\circ C\;({k_f} = 1.86\;K{m^{ - 1}})$ the degree of ionisation of the electrolyte is

An organic liquid, $A$ is immiscible with water. When boiled together with water, the boiling point is $90^{\circ} \mathrm{C}$ at which the partial pressure of water is $526 \mathrm{~mm} \mathrm{Hg}$. The atmospheric pressure is $736 \mathrm{~mm} \mathrm{Hg}$. The weight ratio of the liquid and water collected is $2.5: 1$. The molecular weight of the liquid is ...

The boiling point of water $\left(100^{\circ} {C}\right)$ becomes $100.52^{\circ} {C}$, if $3 \,grams$ of a nonvolatile solute is dissolved in $200 \,{ml}$ of water. The molecular weight of solute is ( $K_b$ for water is $0.6 {~K}-{m}$ )

$0.010 \mathrm{M}$ solution an acid $H A$ freezes at $-0.0205^{\circ} \mathrm{C}$. If $\mathrm{K}_{\mathrm{f}}$ for water is $1.860 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, the ionization constant of the conjugate base of the acid will be (assume $0.010 \mathrm{M}=0.010 \mathrm{~m}$ )

A solution is made by mixing 300 ml. of 1.5 M $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ + 300 ml. of 2 M $CaS{{O}_{4}}$ + 400 ml. of 3.5 M $CaC{{l}_{2}}.$ Find final molarity of

(a) ${SO}_4^{-2}\quad$ (b) ${Ca}^{2+}$

[Assume complete dissociation of these compounds]

Molal depression constant for a solvent is $4.0\mathrm{ }K\mathrm{ }\,\,kg\mathrm{ }\,\,mo{l^{ - 1}}$. The depression in the freezing point of the solvent for $0.03\mathrm{ }\,\,mol\,\,\mathrm{ k}{\mathrm{g}^{ - 1}}$ solution of ${K_2}S{O_4}$ is (Assume complete dissociation of the electrolyte)

At $35^\circ \mathrm{C}$, the vapour pressure of $C{S_2}$ is 512 mm Hg and that of acetone is 344 mm Hg. A solution of $C{S_2}$ in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the following is

A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass $= 60\;g\;\mathrm{mo}{\mathrm{l}^{ - 1}}$) in the same solvent. If the densities of both the solutions are assumed to be equal to $\mathrm{1}\mathrm{.0}\;\mathrm{g}\;\mathrm{c}{\mathrm{m}^{ - 3}},$ molarmass of the substance will be

A solution has $80 \% \frac{\mathrm{w}}{\mathrm{w}} \mathrm{NaOH}$ with density $2 \text{g L}^{-1}.$ Find

(a) Molarity $\quad$ (b) Molality of solution

A solution is prepared by mixing $8.5 \mathrm{~g}_{\text {of }} \mathrm{CH}_2 \mathrm{Cl}_2$ and $11.95 \mathrm{~g}$ of $\mathrm{CHCl}_3$. If vapour pressure of $\mathrm{CH}_2 \mathrm{Cl}_2$ and $\mathrm{CHCl}_3$ at $298 \mathrm{~K}$ are 415 and $200 \mathrm{~mmHg}$ respectively, the mole fraction of $\mathrm{CHCl}_3$ in vapour form is :
(Molar mass of $\mathrm{Cl}=35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )

What volumes of $1 \mathrm{~M} \, \, \& \, \, 2 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4$ solution are required to produce $2 \mathrm{~L}$ of $1.75 \mathrm{~M~H}_2 \mathrm{SO}_4$ solution?

The degree of dissociation $(\alpha)$ of a week electrolyte, $\mathrm{A}_2 \mathrm{SO}_4$ is related to Vant Hoff factor (i) by the expression

$1.2 \% \mathrm{NaCl}$ solution is isotonic with $7.2 \%$ glucose solution. What will be the van't Hoff factor ' $i$ ' for $\mathrm{NaCl}$ ?

The relationship which describes the variation of vapour pressure of a liquid with temperature is called :

A current of dry air was passed through a series of bulbs containing $1.25 \mathrm{~g}$ of a solute $\mathrm{A}_2 \mathrm{B}$ (molar mass $=90 \mathrm{~g} / \mathrm{mol}$) in $49 \mathrm{~g}$ of water and then through pure water. The loss in weight of the former series of bulbs was $0.98 \mathrm{~g}$ and in the later series $0.01 \mathrm{~g}.$ The percentage dissociation of solute is

Acetic acid exists in benzene solution in the dimeric form. In an actual experiment the Van't Hoff factor of acetic acid in benzene solution was found to be 0.52 . Calculate the degree of association and osmotic pressure of $0.1 \mathrm{M}$ acetic acid solution in benzene at $300 \mathrm{~K}$

Dry air was passed successively through a solution containing 5 g of solute (non-volatile and non-electrolytic) in 90 ml $\mathrm{H}_2 \mathrm{O}$ and then through pure $\mathrm{H}_2 \mathrm{O}.$ The loss in weight of solution was 2 g and that of pure solvent 0.05 g. The mol. wt. of solute (in gm) is: (Given: density of water = 1 g/cc)

Assertion: Isotonic solutions don't show net phenomenon of osmosis.

Reason: Isotonic solutions have equal osmotic pressure at constant temperature.

Calculate osmotic pressure of a solution obtained by mixing $100 \mathrm{~mL}$ of $3.4 \%$ (mass/vol.) solution of urea (molar mass 60 ) and $100 \mathrm{~mL}$ of $1.6 \%$ solution of cane-sugar (molar mass 342 ) at $20^{\circ} \mathrm{C}$.

The partition coefficient of solute $X$ in between immiscible liquids $A$ and $B$ is 10 in favour of $A$. The partition coefficient of $X$ in favour of $B$ is :

Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300 K when mixed in the molar ratio of 1:1 and a vapour pressure of 350 mm when mixed in the molar ratio of 1:2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are

Pure benzene freezes at $5.3^{\circ} \mathrm{C}$. A solution of $0.223 \mathrm{~g}$ of phenylacetic acid $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{COOH}\right)$ in $4.4 \mathrm{~g}$ of benzene $\left(K_f=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$ freezes at $4.47^{\circ} \mathrm{C}$. From this observation, one can conclude that:

Assertion (A) : Addition of $\mathrm{HgI}_2$ to $\mathrm{KI}(a q)$ shows an increase in freezing point.

Reason (R) : $\mathrm{HgI}_2$ (insoluble) reacts with $\mathrm{KI}(a q)$ to form complex $\mathrm{K}_2 \mathrm{HgI}_4$ and thus, number of particles present in solution decreases.

A solution containing $30 \,{gms}$ of non-volatile solute in exactly $90 \,{gm}$ water has a vapour pressure of $21.85 {~mm} \,{Hg}$ at $25^{\circ} {C}$. Further $18 \,gms$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 {~mm} \,{Hg}$ at $25^{\circ} {C}$. Calculate the molecular weight of the solute

The vapour pressure of pure benzene and toluene at a particular temperature are $100 {~mm}$ and $50 {~mm}$ respectively. Then the mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene is