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A nonvolatile organic compound ZZ was used to make up two solutions. Solution A\mathrm{A} contains 5.00 g5.00 \mathrm{~g} of Z\mathrm{Z} dissolved in 100 g100 \mathrm{~g} of water, and solution B contains 2.31 g2.31 \mathrm{~g} of ZZ dissolved in 100 g100 \mathrm{~g} of benzene. Solution A\mathrm{A} has a vapor pressure of 754.5mmHg754.5 \mathrm{mmHg} at the normal boiling point of water, and solution B has the same vapor pressure at the normal boiling point of benzene. Calculate the molar mass of Z\mathrm{Z} in solutions A\mathrm{A} and B\mathrm{B} and account for the difference.

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In this exercise, the goal is to determine the molar mass of the compound in the two solutions and compare. Since the problem provided us with the vapor pressures of the solution, we need to use the formula for the vapor-pressure lowering, which is given by the equation:

Pi=xiPi(1)P_\text{i}=x_\text{i}P^*_\text{i} \tag 1

where,

Pi\quad P_\text{i} is the vapor pressure of the solution

xi\quad x_\text{i} is the mole fraction of the solvent

Pi\quad P^*_\text{i} is the vapor pressure of the pure solvent

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