Question

Question 15 of 20 Submit The percent by mass of methanol (MM 32.04 g/mol) in an aqueous solution is 22.3%. What is the molality of the methanol solution? 1 2 3 X 4 5 6 C 7 8 +- x 100 Tap here or pull up for additional resources

Answer 1

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density of solution be same as density of water.

d = 1.0 g/mL

use:

mass = density * volume

= 1 g/mL *1*10^3 mL

= 1*10^3 g

This is mass of solution

mass of CH3OH = 22.3 % of mass of solution

= 22.3*1000.0/100

= 223.0 g

Molar mass of CH3OH = 32.04 g/mol

mass(CH3OH)= 223.0 g

use:

number of mol of CH3OH,

n = mass of CH3OH/molar mass of CH3OH

=(2.23*10^2 g)/(32.04 g/mol)

= 6.96 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 6.96/1

= 6.96 M

Answer: 6.96 M