On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressure of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25 g of heptane and 35 g of octane will be (molar mass of heptane `= 100 g mol^(-1) " and of octane " = 114 g mol^(-1)` )

Options

A. 144.5 kPa
B. 72.0 kPa
C. 36.1 kPa
D. 96.2 kPa

Answer

B

Explanation

Given, `P_("heptane")^@ = 105 kPa`
`P_("octane")^@ = 45 kPa, W_("heptane") = 25 g, W_("octane") = 35 g`
`M_("heptane") =25/100 = 0.25implies N_("octane") =35/114 =0.30`
`x_("heplane") =(0.25)/(0.25+ 0.30) =0.45 implies x_("octane")=(0.30)/(0.25 +0.30)=0.54`
`P_("Total") = X_("heptane") P_("heptane")^@ + X_("octane") P_("octane")^@ = 0.45 xx 105 +0.54 xx 45 = 47.25 + 24.3 = 71.55~~ 72 kPa`