Question

at 40 degree Celsius the vapour pressure of pure liquids benzene and toluene are 160 mm Hg and 60 mm Hg respectively at the same temperature and pressure of an equimolar solution of the two liquids assuming the ideal solution should be

Answer 1

Answer:

110

Explanation:

using the formula

ptotal= p°x1+ px2

as equimolar mixture

X1=0.5 and X2 = 0.5

Answer 2

The question you have asked id incomplete.

I think this is your question:-

At 40°C, the vapour pressure of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg, respectively. At the same temperature, the vapor pressure of an equimolar solution of the two liquids, assuming the ideal solution, should be:

Answer:

110mm of Hg

Explanation:

The question here is to find total pressure exerted by vapour ($P_t$) of both the liquids, which is given as

$=>P_t$=pressure due to toluene +pressure due to benzene

$=>P_t=(P^{0}_b_e_n_z_e_n_e)X_b_e_n_z_e_n_e+(P^{0}_t_o_l_u_e_n_e)X_t_o_l_u_e_n_e$

Since χ = mole fraction = 1/2 [∵ equimolar solutions]

$=>P_t=(160)(\frac{1}{2} )+60(\frac{1}{2} )\\\\=>\boxed{P_t=110}$

Hope this answer helped you :)