What is the molality of 153 mL of a 1.10 M NaF (41.99 g/mol) solution? The density of the solution is 1.23 g/mL.
A)0.894 m
B)0.929 m
C)1.10 m
D)2.34 m
E)5.85 m

Question

Quizplus · Accepted Answer

First, calculate the number of moles of NaF present in the solution:

1.10 moles/L x 0.153 L = 0.1683 moles NaF

Next, convert the mass of the solution to grams:

153 mL x 1.23 g/mL = 188.19 g

Then, calculate the molality:

molality = moles of solute / mass of solvent (in kg)

mass of solvent = mass of solution - mass of solute
mass of solvent = 188.19 g - (0.1683 moles x 41.99 g/mol) = 180.31 g

molality = 0.1683 moles / 0.18031 kg = 0.929 m

Therefore, the best choice is B) 0.929 m.

What Is the Molality of 153 ML of a 1