On mixing, heptane and octane form an ideal solution. At $$373 K,$$  the vapour pressures of the two liquid components (heptane and octane) are $$105$$ $$kPa$$  and $$45$$ $$kPa$$  respectively. Vapour pressure of the solution obtained by mixing $$25.0 g$$  of heptane and $$435 g$$  of octane will be
( molar mass of heptane = $${100\,g\,mo{l^{ - 1}}}$$   and of octane = $${114g\,mo{l^{ - 1}}}$$   )

Solution :
$$\eqalign{ & {P_{{\text{Total}}}} = {P^ \circ }_A{x_A} + {P^ \circ }_B{x_B} \cr & = {P^ \circ }_{{\text{Heptane}}}\,{x_{{\text{Heptane}}}} + {P^ \circ }_{{\text{Octane}}}\,{x_{{\text{Octane}}}} \cr & = 105 \times \frac{{\frac{{25}}{{100}}}}{{\frac{{25}}{{100}} + \frac{{35}}{{114}}}} + 45 \times \frac{{\frac{{35}}{{114}}}}{{\frac{{25}}{{100}} \times \frac{{35}}{{114}}}} \cr & = 105 \times \frac{{0.25}}{{0.25 + 0.3}} + 45 \times \frac{{0.3}}{{0.25 + 0.3}} \cr & = \frac{{105 \times 0.25}}{{0.55}} + \frac{{45 \times 0.3}}{{0.55}} \cr & = \frac{{26.25 + 13.5}}{{0.55}} \cr & = 72\,kPa \cr} $$