A solution of acetone in ethanol shows a positive deviation from Raoult’s law due to miscibility of these two liquids with difference of polarity and length of hydrocarbon chain.
292.
Which one of the following binary mixtures forms an azeotrope with minimum boiling point type?
Minimum boiling azeotrope is formed by solution showing positive deviation. e.g. acetone – ethanol.
293.
The boiling point of $$0.2\,mol\,k{g^{ - 1}}$$ solution of $$X$$ in water is greater than equimolal solution of $$Y$$ in water. Which one of the following statements is true in this case?
A
$$X$$ is undergoing dissociation in water.
B
Molecular mass of $$X$$ is greater than the molecular mass of $$Y.$$
C
Molecular mass of $$X$$ is less than the molecular mass of $$Y.$$
D
$$Y$$ is undergoing dissociation in water while $$X$$ undergoes no change.
Answer :
$$X$$ is undergoing dissociation in water.
Molality of solution $$X=$$ molality of solution $$Y$$
$$ = 0.2\,mol/kg$$
We know that, elevation in the boiling point $$\left( {\Delta {T_b}} \right)$$ of a solution is proportional to the molal concentration of the solution i.e.
$$\Delta {T_b} \propto m\,\,{\text{or}}\,\,\Delta {T_b} = {K_b}m$$
where, $$m$$ is the molality of the solution and $${K_b}$$ is molal boiling point constant or ebullioscopic constant.
∴ By elevation in boiling point relation
$$\Delta {T_b} = i{K_b}m\,\,\,{\text{or}}\,\,\,\Delta {T_b} \propto i$$
where, $$i$$ is van't Hoff factor
Since, $$\Delta {T_b}$$ of solution $$X$$ is greater than $$\Delta {T_b}$$ of solution $$Y.$$
( Observed colligative property is greater than normal colligative property ).
∴ $$i$$ of solution $$X$$ > $$i$$ of solution $$Y$$
∴ Solution $$X$$ undergoing dissociation.
294.
Which one of the following aqueous solutions will exihibit highest boiling point ?
$$\Delta {{\text{T}}_b} = {K_b} \times m$$
Elevation in boiling point is a colligative property, which depends upon the no. of particles ( concentration of solution ). Thus greater the number of particles, greater is the elevation in boiling point and hence greater will be its boiling point.
$$N{a_2}S{O_4} \rightleftharpoons 2N{a^ + } + SO_4^{2 - }$$
Since $$N{a_2}S{O_4}$$ has maximum number of particles hence it has maximum boiling point.
295.
$$1.00\,g$$ of a non-electrolyte solute ( molar mass $$250\,g\,mo{l^{ - 1}}$$ ) was dissolved in $$51.2\,g$$ of benzene. If the freezing point depression constant, $${k_f}$$ of benzene is $$5.12\,K\,kg\,mo{l^{ - 1}},$$ the freezing point of benzene will be lowered by
In ideal solution there is no change in enthalpy and volume after mixing, i.e.
$$\eqalign{
& \Delta {V_{{\text{mixing}}}} = 0,\,\Delta {H_{{\text{mixing}}}} = 0 \cr
& {\text{and}}\,\,\,{T_{{\text{mix}}}} = 0 \cr} $$