Question

The vapor pressure of ether (molecular mass=74) is 442 mm Hg at 293 K. If 3g of a compound A are dissolved in 50 g of ether at this temperature, the vapor pressure falls to 426 mm Hg. Calculate the molecular mass of A assuming that the solution of A is very dilute.
A. $121\;{\rm{gmo}}{{\rm{l}}^{ - 1}}$
B. $122\;{\rm{gmo}}{{\rm{l}}^{ - 1}}$
C. $123\;{\rm{gmo}}{{\rm{l}}^{ - 1}}$
D. $124\;{\rm{gmo}}{{\rm{l}}^{ - 1}}$

Answer 1

We know that colligative properties are the properties which depend on the number of solute particles to the total number of particles in the solution not considering the solute's nature.

Complete answer:
If a solution contains non-volatile solute, the lowering of vapor pressure depends on the sum of mole fraction of different solutes. So, the mathematical equation representing the condition is,

$\dfrac{\Delta p_1}{{p_1}^0}=x_2$
$\dfrac{{p_1}^0- p_1}{{p_1}^0}=x_2$ …… (1)

Where, $\Delta {p_1}$ is lowering of vapor pressure, $p_1^0$ is the vapor pressure of pure solvent, ${p_1}$ of solution and ${x_2}$ is mole fraction of solute.

Now, come to the question. Given that vapor pressure of ether is 442 mm Hg. When a solute A is added to ether, vapor pressure falls to 426 mm Hg. So, $p_1^0 = 442\;{\rm{mm}}\;{\rm{Hg}}$ and ${p_1} = 426\;{\rm{mm}}\;{\rm{Hg}}$

Substitute the value of $p_1^0$ and ${p_1}$ in equation (1).

$\dfrac{{442\;{\rm{mm}}\;{\rm{Hg}} - 426\;{\rm{mm}}\;{\rm{Hg}}}}{{442}} = {x_A}$ ….. (2)

Now, write the mole fraction part of the above equation. Here, solute A of 3 g is added to 50 g ether. So, mole fraction of the solute A is,

${x_A} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{A}}}}{{{\rm{Moles}}\;{\rm{of}}\;{\rm{A}} + {\rm{Moles}}\;{\rm{of}}\;{\rm{ether}}}}$ …… (3)

As solution of A is dilute, moles of A <<< moles of ether

So, ${\rm{Moles\;of\;A}} + {\rm{ Moles\;of\;ether}} \approx {\rm{Moles\;of\; ether}}$

So, equation (3) becomes,

${x_A} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{A}}}}{{{\rm{Moles}}\;{\rm{of}}\;{\rm{ether}}}}$

We know that moles can be calculated by dividing mass by molar mass. So, the above equation becomes,

${x_A} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{A}}}}{{\dfrac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{ether}}}}{{{\rm{Molecular}}\,{\rm{Mass}}\;{\rm{of}}\;{\rm{ether}}}}}}$

Now, we substitute ${x_A}$ in equation (2).

$\dfrac{{442\;{\rm{mm}}\;{\rm{Hg}} - 426\;{\rm{mm}}\;{\rm{Hg}}}}{{442\;{\rm{mm}}\;{\rm{Hg}}}} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{A}}}}{{\dfrac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{ether}}}}{{{\rm{Molecular}}\,{\rm{Mass}}\;{\rm{of}}\;{\rm{ether}}}}}}$

Now, put the mass of ether 50 g and molecular mass of ether $74\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$in the above equation and solve for moles of A.

$\begin{array}{c}\dfrac{{442\;{\rm{mm}}\;{\rm{Hg}} - 426\;{\rm{mm}}\;{\rm{Hg}}}}{{442\;{\rm{mm}}\;{\rm{Hg}}}} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{A}}}}{{\dfrac{{50\;{\rm{g}}}}{{74\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}}}\\\\\dfrac{{16}}{{442}} = \dfrac{{{\rm{Moles}}\;{\rm{of}}\;{\rm{A}}}}{{0.6756\;mol}}\end{array}$

Now, calculate moles of A.

$\begin{array}{c}{\rm{Moles}}\;{\rm{of A}} = \dfrac{{16 \times 0.6756}}{{442}}{\rm{mol}}\\\\{\rm{ = 0}}{\rm{.0244}}\;{\rm{mol}}\end{array}$

So, moles of A in the solution is 0.0244 mol.

Now, we use the formula of moles to calculate the molecular mass of A.

${\rm{Number}}\;{\rm{of}}\;{\rm{moles = }}\dfrac{{{\rm{Mass}}}}{{{\rm{Molar \,Mass}}}}$

We rearrange the above formula and put the value of moles of A and mass of A.

$\begin{array}{c}{\rm{Molar \,Mass\, of\, A}} = \dfrac{{{\rm{Mass}}\;{\rm{of}}\,{\rm{A}}}}{{{\rm{Moles}}\;{\rm{of}}\;{\rm{A}}}}\\\ = \dfrac{{3\;{\rm{g}}}}{{0.0244\;{\rm{mol}}}}\\\ = 122.95\,{\rm{gmo}}{{\rm{l}}^{ - 1}}\\\ = 123\;g\,{\rm{mo}}{{\rm{l}}^{ - 1}}\end{array}$

Therefore, molecular mass of A is $123\;g\,{\rm{mo}}{{\rm{l}}^{ - 1}}$. Hence, option C is correct.

Note: It is to be noted that, for dilute solution the moles of solute is neglected in the denominator of the lowering of vapour pressure equation. So, the equation becomes $\dfrac{{p_1^0 - {p_1}}}{{p_1^0}} = \dfrac{{{n_2}}}{{{n_1}}}$, where ${n_2}$ is moles of solute and ${n_1}$ moles of solvent.