Question

A solution is prepared by dissolving 128 g of naphthalene (C10H8) in 780 g of benzene (C6H6). The vapor pressure of pure benzene is 12.6 kPa at 25\(^\circ\)C. Assuming that naphthalene in benzene is an ideal solution, the partial vapor pressure of benzene is _____ kPa (rounded off to two decimal places). 
 

Hint:

For ideal solutions: \(p_i=x_i P_i^\ast\) (Raoult's law).
Nonvolatile solutes (like naphthalene here) lower the solvent vapor pressure in proportion to the solvent mole fraction.

Answer 1

Correct Answer: 11.34

Step 1: Calculate moles.
Molar masses: \(M_{\mathrm{benz}} \approx 78.11~\mathrm{g\,mol^{-1}}\), \(M_{\mathrm{naph}} \approx 128.17~\mathrm{g\,mol^{-1}}\).
\[ n_{\mathrm{benz}}=\frac{780}{78.11}\approx 9.986~\mathrm{mol}, \qquad n_{\mathrm{naph}}=\frac{128}{128.17}\approx 0.999~\mathrm{mol}. \] Step 2: Mole fraction of benzene.
\[ x_{\mathrm{benz}}=\frac{n_{\mathrm{benz}}}{n_{\mathrm{benz}}+n_{\mathrm{naph}}} =\frac{9.986}{9.986+0.999}\approx 0.9091. \] Step 3: Apply Raoult's law.
\[ p_{\mathrm{benz}}=x_{\mathrm{benz}}\,P^{\ast}_{\mathrm{benz}} =0.9091\times 12.6~\mathrm{kPa}\approx 11.454~\mathrm{kPa} \Rightarrow \boxed{11.45~\mathrm{kPa}} \text{ (to two decimals)}. \]