The number of moles of chlorobenzene and bromobenzene are $0.1$ and $0.2$ respectively. The vapour pressure of pure chlorobenzene and pure bromobenzene are $0.350 \ bar$ and $0.500 \ bar$ respectively. Calculate the total vapour pressure of the solution prepared by mixing these two components.
(0.450 BAR) According to Raoult's law,the total vapour pressure $P_{total} = P_A + P_B = x_A P_A^0 + x_B P_B^0$.
Given:
Moles of chlorobenzene $(n_A)$ = $0.1$
Moles of bromobenzene $(n_B)$ = $0.2$
Total moles = $0.1 + 0.2 = 0.3$
Mole fraction of chlorobenzene $(x_A)$ = $0.1 / 0.3 = 1/3$
Mole fraction of bromobenzene $(x_B)$ = $0.2 / 0.3 = 2/3$
$P_A^0 = 0.350 \ bar$
$P_B^0 = 0.500 \ bar$
$P_{total} = (1/3 \times 0.350) + (2/3 \times 0.500)$
$P_{total} = 0.1167 + 0.3333 = 0.450 \ bar$.
Given:
Moles of chlorobenzene $(n_A)$ = $0.1$
Moles of bromobenzene $(n_B)$ = $0.2$
Total moles = $0.1 + 0.2 = 0.3$
Mole fraction of chlorobenzene $(x_A)$ = $0.1 / 0.3 = 1/3$
Mole fraction of bromobenzene $(x_B)$ = $0.2 / 0.3 = 2/3$
$P_A^0 = 0.350 \ bar$
$P_B^0 = 0.500 \ bar$
$P_{total} = (1/3 \times 0.350) + (2/3 \times 0.500)$
$P_{total} = 0.1167 + 0.3333 = 0.450 \ bar$.
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