Question

solution of the two ?
7. The vapour pressure of ethyl acetate and ethyl propionate are 72.8 and 27.7 mm of Hg respectively. A solution
is prepared by mixing 25 g ethyl acetate and 50 g ethyl propionate. Assuming the solution to be ideal,
calculate the vapour pressure of the solution.
8. Benzene and toluene form nearly ideal solution. At a certain temperature, the vapour pressure of the pure
TYC
H
2.
Dobia ambania malalata tha vananenmassima​

Answer 1

Ans 7. The vapour pressure of the solution is 44.25 mm of Hg.

Ans. 8 The vapour pressure of the solution containing equal moles of two substances is 102.5 mm of Hg

Explanation:

Ans 7.

The vapour pressure of ethyl acetate = 72.8

The vapour pressure of ethyl propionate = 27.7 mm

=> Mole fraction of ethyl acetate and ethyl propionate:

molefraction of ethyl acetate, x_e.a. = 25/88 / (25/88 + 50/102) = 0.367

molefraction of ethyl propionate,  x_e.p. = 50/102 / (25/88 + 50/102) = 0.633

=> The vapour pressure of the solution, Ptotal:

Ptotal = Pe.a. * x_e.a. + Pe.p. * x_e.p.

= 72.8 * 0.367 + 27.7 * 0.633

= 26.72 + 17.53

= 44.25 mm of Hg

Thus, the vapour pressure of the solution is 44.25 mm of Hg.

Q:8 Benzene and toluene form nearly ideal solution. At a certain temparature, calculate the vapour pressure of the solution containing equal moles of two substances.

=>  Here, pure partial pressure of benzene=150 mm of Hg,

pure partial pressure of toluene=55 mm of Hg

moles of benzene = 0.5

moles of toluene = 0.5

=> the vapour pressure of the solution of benzene and toluene:

Ptotal = 150 * 0.5 + 55 * 0.5

          = 75 + 27.5

          = 102.5 mm of Hg

Thus, the vapour pressure of the solution containing equal moles of two substances is 102.5 mm of Hg .

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