Total vapour pressure of a mixture of 1 , mol | vedclass

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Question

(A) According to Raoult's law,the ideal total vapour pressure $(P_{total, ideal})$ is given by: $P_{total, ideal} = X_x P_x^o + X_y P_y^o$ Given: $n_x

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There is a negative deviation from Raoult's law

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(A) According to Raoult's law,the ideal total vapour pressure $(P_{total, ideal})$ is given by: $P_{total, ideal} = X_x P_x^o + X_y P_y^o$ Given: $n_x = 1 \, mol$,$n_y = 2 \, mol$. Total moles = $1 + 2 = 3 \, mol$. Mole fractions: $X_x = 1/3$,$X_y = 2/3$. $P_{total, ideal} = (1/3 	imes 150) + (2/3 	imes 300) = 50 + 200 = 250 \, torr$. The observed total vapour pressure is $240 \, torr$. Since $P_{observed} < P_{total, ideal}$ $(240 \, torr < 250 \, torr)$,the mixture shows a negative deviation from Raoult's law.

Total vapour pressure of a mixture of $1 \, mol \, X$ $(P_x^o = 150 \, torr)$ and $2 \, mol \, Y$ $(P_y^o = 300 \, torr)$ is $240 \, torr$. In this case:

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