At 300 K,the vapour pressure of an ideal sol | vedclass

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(C) For the initial solution: $n_A = 3, n_B = 2$. Total moles = $5$. $P_1 = p_A^o \left( \frac{3}{5} \right) + p_B^o \left( \frac{2}{5} \right) = 600

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) For the initial solution: $n_A = 3, n_B = 2$. Total moles = $5$. $P_1 = p_A^o \left( \frac{3}{5} \right) + p_B^o \left( \frac{2}{5} \right) = 600 \ torr$. $3p_A^o + 2p_B^o = 3000$ (Equation $1$). After adding $1.5 \ mol$ of $A$ and $0.5 \ mol$ of $C$: $n_A = 4.5, n_B = 2, n_C = 0.5$. Total moles = $7$. Since $C$ is non-volatile,$p_C^o = 0$. $P_2 = 600 + 30 = 630 \ torr$. $P_2 = p_A^o \left( \frac{4.5}{7} \right) + p_B^o \left( \frac{2}{7} \right) = 630$. $4.5p_A^o + 2p_B^o = 4410$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(4.5 - 3)p_A^o = 4410 - 3000 \implies 1.5p_A^o = 1410 \implies p_A^o = 940 \ torr$. Substituting $p_A^o$ in Equation $1$: $3(940) + 2p_B^o = 3000 \implies 2820 + 2p_B^o = 3000 \implies 2p_B^o = 180 \implies p_B^o = 90 \ torr$.

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