A solution is prepared by dissolving 4.21 g of a nonelectrolyte in 50.0 g of water.If the boiling point increases by 0.203°C,what is the molar mass of the solute? The boiling point elevation constant for water is 0.512°C/m.

Question

Quizplus · Accepted Answer

First, we can calculate the molality (b) of the solution using the formula:

ΔT = Kb * b

where ΔT is the boiling point elevation, Kb is the boiling point elevation constant for water, and b is the molality of the solution.

Substituting the given values, we get:

0.203°C = 0.512°C/m * b

b = 0.3975 m

Next, we can calculate the moles of the solute using the formula:

moles solute = m * kg solvent / M

where m is the molality of the solution, kg solvent is the mass of the solvent (in kg), and M is the molar mass of the solute.

Substituting the given values, we get:

moles solute = 0.3975 * 0.05 / M

moles solute = 0.019875 / M

Finally, we can use the mass of the solute and the moles of the solute to calculate the molar mass of the solute:

molar mass = mass solute / moles solute

molar mass = 4.21 g / 0.019875 mol

molar mass = 211.4 g/mol

Rounding off to the nearest whole number, we get:

molar mass = 212 g/mol

Therefore, the correct answer is D) 212 g/mol.

A Solution Is Prepared by Dissolving 4