Solutions practice · Mixed difficulty

Statement I: The freezing point of water is depressed by the addition of glucose.

Statement II: Entropy of solution is less than entropy of pure solvent.

Consider following terms $(\mathrm{m}=$ molality $)$ :

I : ${mK}_{{b}};\quad$ II: ${mK}_{{b}} {i}\quad$ III : $\frac{\Delta {T}_{{b}}}{{i}}\quad$ IV : $K_b$

Terms which can be expressed in degree (temperature) are

The degree of dissociation ($\propto$) of a weak electrolyte, ${A_x}By$ is related to vant Hoff factor (i) by the expression

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

50 g of an acid is dissolved in one litre aqueous solution. Distribution coefficient in favour of ether is 3. Acid left in aqueous layer when solution is shaken with one litre ether :

The solution from the following with highest depression in freezing point/lowest freezing point is

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 : On adding $\mathrm{NaCl}$ to water les vapour piessure imcieases.

REASON : Addition of non-volatile solute increases the vapour pressure.

The osmotic pressure of a non-volatile solute in $\mathrm{C}_6 {H}_6$ at $25^{\circ} {C}$ is $20.66 {Nm}^{-2}$. If the solution had a concentration of $2 {~g} / {dm}^3$, what is molar mass of solute?

Select correct statement(s) :

5 L of a solution contains 25 mg of $CaC{O_3}.$ What is its concentration in ppm? (mol. wt. of $CaC{O_3}$ is 100)

When ${Hgl}_2$ is added to an aqueous solution of ${Kl}$

During depression of freezing point in a solution the following are in equilibrium

$0.1435 \mathrm{~m}$ solution of a non-volatile, non-electrolyte solute has the freezing point 0.73 degrees lower than that of benzene. What is the value of molal freezing point depression constant of benzene?

ASSERTION : Molar heat of vaporisation of water is greater than benzene.

REASON : Molar heat of vaporisation is the amount of heat required to vaporise one mole of liquid at constant temperature.

The plots of $\frac{1}{{{X_A}}}\;vs.\;\frac{1}{{{Y_A}}}$ (where ${X_A}\;\mathrm{and}\;{Y_A}$ are the mole fraction of liquid A in liquid and vapour phase respectively) is linear with slope and intercepts respectively:

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.

The ratio of vapour pressure over solution phase on mixing two immiscible liquids is equal to :

One gram of silver gets distributed between $10{\kern 1pt} \,\mathrm{c}{\mathrm{m}^3}$ of molten zinc and $100\,\mathrm{c}{\mathrm{m}^3}$ of molten lead at $8000^\circ C.$ The percentage of silver still left in the lead layer in approximately

18 g of Glucose is added to 178.2 g of water the vapour pressure of water for this aqeous solution at $100^\circ C$ is.

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

According to phase rule, if $P = 3,C = 1,$ then F must be equal to :

Aqueous solutions of two acids having same concentrations of common ion $\left[ {{\mathrm{H}^ + }} \right]$ are called:

Assertion (A): Sum of mole fraction of all components in a mixture is 1.

Reason (R): Mole fraction is temperature dependent mode of concentration.

The osmatic pressure of a 5% (wt./vol) solution of cane sugar at $150^\circ C$ is

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

Assertion (A) : The pressure exerted by a liquid in equilibrium with its liquid at a given temperature is called its vapour pressure.

Reason (R) : If a non-volatile solute is added to a solvent to give a solution, the vapour pressure of the solution is found to be greater than the vapour pressure of the pure solvent.

Two elements $A$ and $B$ form compounds having formulae $A B_2$ and $A B_4$. When dissolved in $20 \mathrm{~g}$ benzene, $1 \mathrm{~g}$ of $A B_2$ lowers the freezing point by $2.3^{\circ}$ whereas $1.0 \mathrm{~g}$ of $A B_4$ lowers it by $1.3^{\circ} \mathrm{C}$. $K_f$ for benzene is $5.1 \mathrm{~K} \mathrm{~mol}^{-1} \mathrm{~kg}$. The atomic masses of $A$ and $B$ are respectively

A liquid is in equilibrium with its vapours at its boiling point. On the average the molecules in the two phases have equal :

Which of the following is incorrect regarding solution showing negative deviation from Raoult's law?