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Question

Benzene and toluene form an ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $$300\ K$$ are $$50.71\ mm\ Hg$$ and $$32.06\ mm\ Hg$$ respectively. Calculate the mole fraction of benzene in vapour phase if $$80\ g$$ of benzene is mixed with $$100\ g$$ of toluene.


Solution

Number of moles is the ratio of mass to molar mass.

The molar masses of benzene and toluene are $$78$$ g/mol and $$92$$ g/mol respectively.

Number of moles of benzene $$\displaystyle = \dfrac {80}{78} = 1.026$$

Number of moles of toluene $$\displaystyle = \dfrac {100}{92} = 1.087$$

Mole fraction of benzene, $$\displaystyle X_B = \dfrac {1.026}{1.026+1.087}=0.486$$

Mole fraction of toluene, $$\displaystyle X_T = 1-0.486 = 0.514$$

$$\displaystyle P_B = P^o_B \times X_B= 50.71 \times 0.486=24.65$$ mm Hg

$$\displaystyle P_T = P^o_TX_T = 32.06 \times 0.514 = 16.48$$ mm of Hg

Total vapour pressure $$\displaystyle =24.65+16.48=41.13$$ mm Hg

Mole fraction of benzene in vapour phase is as follows:
$$\displaystyle Y_B = \dfrac {24.65}{41.13} = 0.60$$

Chemistry
NCERT
Standard XII

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