Question

the vapour pressure of ether (mol mass = 74) is 442 mm hg at 293 k. if 3g of a compound b are dissolved in 50 g of ether at this temperature, the vapour pressure falls to 426 mm hg. calculate the molecular mass of b. assume that the solution of b in ether is very dilute.
in this case: $w_b$, the mass of solute (b) = 3 g
$w_a$, the mass of solvent (ether) = 50 g,
$m_b$, the mol mass of solute b =?
$m_a$, the mol mass of solvent (ether) = 74,
$p$, the vapour pressure of solvent (ether) = 442 mm,
$p_s$, the vapour pressure of solution = 426 mm

Explanation:

Step1: Apply Raoult's law for dilute solutions

For a dilute solution, $\frac{P - P_s}{P}=\frac{n_B}{n_A + n_B}\approx\frac{n_B}{n_A}$, where $P$ is the vapour - pressure of the pure solvent, $P_s$ is the vapour - pressure of the solution, $n_A$ is the number of moles of the solvent and $n_B$ is the number of moles of the solute.
$n_A=\frac{w_A}{M_A}$ and $n_B=\frac{w_B}{M_B}$, so $\frac{P - P_s}{P}=\frac{\frac{w_B}{M_B}}{\frac{w_A}{M_A}}$.

Step2: Substitute the given values

We know that $P = 442$ mm Hg, $P_s=426$ mm Hg, $w_B = 3$ g, $w_A = 50$ g and $M_A = 74$ g/mol.
First, calculate $\frac{P - P_s}{P}=\frac{442 - 426}{442}=\frac{16}{442}$.
And $\frac{\frac{w_B}{M_B}}{\frac{w_A}{M_A}}=\frac{w_BM_A}{w_AM_B}$. Substituting the values, we get $\frac{16}{442}=\frac{3\times74}{50\times M_B}$.

Step3: Solve for $M_B$

Cross - multiply the equation $\frac{16}{442}=\frac{3\times74}{50\times M_B}$ to get $16\times50\times M_B=3\times74\times442$.
$M_B=\frac{3\times74\times442}{16\times50}$.
$M_B=\frac{3\times74\times442}{800}$.
$M_B=\frac{3\times74\times221}{400}$.
$M_B=\frac{3\times16354}{400}=\frac{49062}{400}=122.655\approx123$ g/mol.

Answer:

123 g/mol