A very small amount of non-volatile solute (that does not dissociate) is dissolved in 56.8 cm3 of be

Question

A very small amount of non-volatile solute (that does not dissociate) is dissolved in 56.8 cm3 of benzene (density 0.889 g cm−3). At room temperature, vapour pressure of this solution is 100 mm Hg while that of benzene is 102 mm Hg. If the freezing temperature of this solution is 1.3 degree lower than that of benzene, then what is the value of molal freezing point depression constant of benzene?

Accepted Answer

If the molarity of solution is m, then$\frac{P^{o} - P}{P} = \frac{n_{1}}{n_{2}} \Rightarrow \frac{2}{100} = \frac{m}{1000/78} \Rightarrow m = \frac{10}{39}$Now, $\Delta T_{f} = K_{f}.m \Rightarrow 1.3 = K_{f} 	imes \frac{10}{39} \Rightarrow K_{f} = 5.07	ext{ K/m}$

Question

Options

Solution

More Solution Questions