2.18    Calculate the mass of a non-volatile solute (molar \; mass\; 40 \; g \; mol^{-1})  which should be dissolved in    114\; g  octane to reduce its vapour pressure to  80\%.

Answers (1)

Let the initial vapour pressure of octane = p_o^{\circ}.

After adding solute to octane, the vapour pressure becomes :

                                                                     =\frac{80}{100}\times p_o^{\circ} = 0.8p_o^{\circ}

Moles of octane :

                                    = \frac{114}{114} = 1               \left ( Molar\ mass\ of\ octane = 8(12) + 18(1) = 114\ g\ mol^{-1} \right )

Using Raoult's law we get :

                                           \frac{p_o^{\circ} - p}{p_o^{\circ}} = x_2

or                                     \frac{p_o^{\circ} -0.8p_o^{\circ} }{p_o^{\circ}} = \frac{\frac{W}{40}}{\frac{W}{40}+1}

or                                            w = 10\ g

Thus required mass of non-volatile solute = 10g.

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