Answer :
To find the total vapor pressure of the solution, we can use Raoult's Law, which states that the vapor pressure of a solution is the sum of the partial pressures of its components. According to Raoult's Law:
\[ P_{\text{total}} = P_A^\circ x_A + P_B^\circ x_B \]
where:
- \( P_{\text{total}} \) is the total vapor pressure of the solution,
- \( P_A^\circ \) and \( P_B^\circ \) are the vapor pressures of pure liquids A and B respectively,
- \( x_A \) and \( x_B \) are the mole fractions of A and B in the solution.
First, let's calculate the mole fractions of A and B:
\[ x_A = \frac{{\text{moles of A}}}{{\text{total moles}}} = \frac{2}{2+3} = \frac{2}{5} \]
\[ x_B = \frac{{\text{moles of B}}}{{\text{total moles}}} = \frac{3}{2+3} = \frac{3}{5} \]
Now, we can substitute the given values into Raoult's Law:
\[ P_{\text{total}} = (100 \, \text{torr}) \times \left( \frac{2}{5} \right) + (80 \, \text{torr}) \times \left( \frac{3}{5} \right) \]
\[ P_{\text{total}} = \left( 40 \, \text{torr} \right) + \left( 48 \, \text{torr} \right) \]
\[ P_{\text{total}} = 88 \, \text{torr} \]
So, the total pressure of the solution obtained by mixing 2 moles of A and 3 moles of B would be \( 88 \, \text{torr} \), which corresponds to option (C).
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