Question

(a) State Raoult’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law?
(b) 1.00g of non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40K. Find the molar mass of the solute. (${K_f}$for benzene$ = 5.12K\,mol/kg$).

Answer 1

Hint: In this question, we need to state the Raoult’s law for a solution containing volatile components and how the Raoult’s Law is a special case of Henry’s Law. For this, we will first define Raoult's law and then establish the relation between Henry’s constant and Raoult’s law. Moreover, we need to calculate the molar mass of the solute such that $1.00g$ of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40K. For this, we will use the relation of Raoult’s Law.

Complete step by step solution: (a)
According to Raoult’s law the partial vapor pressure in a solution or mixture is directly proportional to its mole fraction present in the solution.
Mathematically, we can represent as ${P_{solution}}\,\propto \,{x_{solvent}}$
Where, ${P_{solution}}$ is the vapour pressure of the solution
${x_{solvent}}$ is the mole fraction of the solvent
i.e., ${P_{solution}} = P_{solvent}^0 \times {x_{solvent}}$ where, $P_{solvent}^0$ is the vapour pressure of the pure solvent.
Now,
Let ${P_1}$ and ${P_2}$ be the partial vapour pressures of two volatile components 1 and 2 respectively. Also, $P_1^0$ and $P_2^0$ are the vapour pressures of pure components. ${x_1}$ and ${x_2}$ are the mole fractions of component 1 and 2.
Applying the Raoult’s law to both component 1 and 2. We get,
${P_1}\,\propto \,{x_2}$ and ${P_2}\,\propto \,{x_2}$
${P_1} = P_1^0 \times {x_1}$ and${P_2} = P_2^0 \times {x_2}$ ---(1)
Now according to Dalton’s law, the total pressure of a mixture is equal to the sum of the partial pressures of its individual gases in the mixture.
i.e., ${P_{total}} = {P_1} + {P_2}$ ---(2)
Applying the equation (1) in (2), we get ${P_{total}} = P_1^0{x_1} + P_2^0{x_2}$
We know that the total mole fraction of all the components will be equal to 1
i.e.,
$
  {x_1} + {x_2} = 1 \\
   \Rightarrow {x_1} = 1 - {x_2} \\
 $
Applying this to${P_{total}}$, we get
${P_{total}} = (1 - {x_2})P_1^0 + P_2^0{x_2}$
Raoult’s law became a special case of Henry’s law.
According to Henry’s law the partial pressure of the gas in a liquid is directly proportional to its mole fraction. Mathematically, $p = {K_H}x$
Where,
$p$ is the partial pressure of the gas
${K_H}$Henry’s law constant
$x$is the mole fraction
When compared with Raoult’s law only the proportionality constant differs. Hence, Raoult’s law becomes a special case of Henry’s law.
(b)
We know that, from the colligative properties of a solvent
${T_f} = {K_f} \times m$ --- (1)
Where ${T_f}$ is the depression in freezing point
${K_f}$is the freezing point depression constant
$m$is the molality
Given that, ${T_f} = 0.40K$ and ${K_f} = 5.12{\text{ kg/mol}}$
From the equation (1)
$m = \dfrac{{{T_f}}}{{{K_f}}}$
Substituting the values in the above equation, we get
$
  m = \dfrac{{0.40}}{{5.12}}{\text{ }}mol\,k{g^{ - 1}} \\
   \Rightarrow m = 0.078{\text{ }}mol\,k{g^{ - 1}} \\
 $
Now by definition of molality,
\[
molality,\,m = \dfrac{{number\,of\,moles\,in\,solute}}{{mass\,of\,solvent\left( {kg} \right)}}
\]
\[
number\,of\,moles\,in\,solute = \dfrac{{mass\,of\,solute}}{{molar\,mass\,of\,solute}}
\]
$mass\,of\,solvent = 50g = 50 \times {10^{ - 3}}kg$
Therefore, $m = \dfrac{{mass\,of\,solute}}{{molar\,mass\,of\,solute \times mass\,of\,solvent(kg)}}$
Substituting the values in the above equations:
\[
  0.078 = \dfrac{1}{{molar\,mass\,of\,solute \times 50 \times 10{}^{ - 3}}} \\
   \Rightarrow molar\,mass\,of\,solute = \dfrac{1}{{0.078 \times 50 \times {{10}^{ - 3}}}}kg\,mo{l^{ - 1}} \\
   \Rightarrow molar\,mass\,of\,solute = 256.41\,kg\,mo{l^{ - 1}} \\
 \]
Therefore the molar mass of solute is $256.41\,kg\,mo{l^{ - 1}}$.

Note: Here, in this question, use of Dalton’s law and Henry’s law is very essential as parameters need to be determined from there only. Here, we can directly apply the equation of ${T_f} = {K_f} \times \dfrac{{{w_2} \times 1000}}{{{M_2} \times {w_1}}}$ as we have known with all the parameters present in the equation where,
${T_f}$ is the depression in freezing point
${K_f}$ is freezing point depression constant
${w_2}$ is the mass of solute
${w_1}$ is the mass of solvent
${M_2}$ is the molar mass of solute.