1 , text{mole} of liquid A and 2 , text{mole} | vedclass

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(B) Given: $n_A = 1 \, 	ext{mole}$,$n_B = 2 \, 	ext{mole}$.Total moles $n_T = 1 + 2 = 3 \, 	ext{mole}$.Mole fractions: $x_A = 1/3$,$x_B = 2/3$.Pure

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Showing negative deviation

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(B) Given: $n_A = 1 \, 	ext{mole}$,$n_B = 2 \, 	ext{mole}$.Total moles $n_T = 1 + 2 = 3 \, 	ext{mole}$.Mole fractions: $x_A = 1/3$,$x_B = 2/3$.Pure vapor pressures: $P_A^0 = 45 \, 	ext{torr}$,$P_B^0 = 36 \, 	ext{torr}$.According to Raoult's Law,the expected vapor pressure $P_{	ext{ideal}} = P_A^0 x_A + P_B^0 x_B$.$P_{	ext{ideal}} = (45 	imes 1/3) + (36 	imes 2/3) = 15 + 24 = 39 \, 	ext{torr}$.The observed vapor pressure is $38 \, 	ext{torr}$.Since $P_{	ext{observed}} < P_{	ext{ideal}}$ $(38 < 39)$,the solution shows negative deviation from Raoult's Law.

$1 \, \text{mole}$ of liquid $A$ and $2 \, \text{mole}$ of liquid $B$ form a solution with a vapor pressure of $38 \, \text{torr}$. If the vapor pressures of pure $A$ and pure $B$ are $45 \, \text{torr}$ and $36 \, \text{torr}$ respectively,the solution will be:

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