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The molecules of the solvent are combining, cutting the concentration of the solvent in half. This, therefore, doubles the molar mass of the solute. Example: 1. Suppose the freezing point of a solution of 2.00 g of an unknown molecular substance in 10.00 g of the solvent benzene is measured. If the solution freezes at a temperature of 6.33°C lower than pure benzene itself, calculate the molar mass of the unknown substance. - 6.33/5.12= 1.24m, solve for x- 1.24m= x/0.01 kg, x=0.0124 mol, 2.00 g/0.0124mol=161.29 g- your molar mass - now cut the amount of solvent, 0.01Kg(10.00g), in half 1.24m=x/0.005 kg, x=0.0062 mol, 2.00 g/0.0062mol = 322.581 g, your larger molar mass

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14y ago
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12y ago

The molar mass would increase with the forming of a dimer because the dimer decreases pressure. If you look at the ideal gas equation you will see that pressure is in the denominator, thus leading to an increase in molar mass.

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Q: What effect would there be on a molar mass determination if the solute were to dimerize?
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