Chemistry - Liquid Solution Question with Solution | TestHub

Let $P_A^o$ and $P_B^o$ be the vapor pressures of pure A and B.

Initial state: $n_A=3$ mol, $n_B=2$ mol. Total moles $n_T=5$ mol.

$x_A=3/5=0.6$, $x_B=2/5=0.4$.

$P_{total,1}=x_A{P}_A^o+x_B{P}_B^o=0.6P_A^o+0.4P_B^o=600$ torr (1)

Final state: $n_A^{\prime }=3+1.5=4.5$ mol, $n_B^{\prime }=2$ mol, $n_C=0.5$ mol (non-volatile).

Total moles $n_T^{\prime }=4.5+2+0.5=7$ mol.

$x_A^{\prime }=4.5/7$, $x_B^{\prime }=2/7$.

$P_{total,2}=x_A^{\prime }{P}_A^o+x_B^{\prime }{P}_B^o=(4.5/7)P_A^o+(2/7)P_B^o=600+30=630$ torr (2)

From (1): $0.6P_A^o+0.4P_B^o=600\Rightarrow 3P_A^o+2P_B^o=3000$ (3)

From (2): $(4.5/7)P_A^o+(2/7)P_B^o=630\Rightarrow 4.5P_A^o+2P_B^o=4410$ (4)

Subtract (3) from (4):

$(4.5-3)P_A^o=4410-3000$

$1.5P_A^o=1410$

$P_A^o=1410/1.5=940$ torr

Substitute $P_A^o=940$ into (3):

$3(940)+2P_B^o=3000$

$2820+2P_B^o=3000$

$2P_B^o=180$

$P_B^o=90$ torr

The final answer is $\boxed{90}$.