Answer to Question #317503 in General Chemistry for Mave

Question

2. A solution was produced by dissolving 3.75 g of a nonvolatile solute in 95g of

acetone. The boiling point of pure acetone was observed to be 55.95 ºC, while the boiling

point of the solution was 56.50ºC. If the boiling point for acetone is 1.71ºC/m, what is the

approximate molar mass of solute?

Step 1: First compute the molality of the boiling point equation.

Step 2. Then, from the definition of molality, compute the number of moles

solute, n(solute), in the sample.

Step 3. Solve for the molar mass.

Please solve it like this

STEP1: SOLUTION

STEP2: SOLUTION

STEP3: SOLUTION

THANK YOUUU

Answer 1

deltaT = C_m x K_b

deltaT = 56.5 - 55.95 = 0.55 degrees

C_m = deltaT / K_b = 0.55 / 1.71 = 0.32 m

The formula for molality is m = moles of solute / kilograms of solvent.

Moles of solute = Сm x kilograms of solvent = 0.32 x 0.095 = 0.03 mol

n = m / M

M_solute = m / n = 3.75 / 0.03 = 125 g/mol