Question

A solution contains 48.6 g glucose (C6H12O6) dissolved in 0.800 L of water. What is the molaltiy of the solution given water has a density of 1 g/mL

Answer 1

Molar mass of C6H12O6,
MM = 6*MM(C) + 12*MM(H) + 6*MM(O)
= 6*12.01 + 12*1.008 + 6*16.0
= 180.156 g/mol


mass(C6H12O6)= 48.6 g

number of mol of C6H12O6,
n = mass of C6H12O6/molar mass of C6H12O6
=(48.6 g)/(180.156 g/mol)
= 0.2698 mol

voluem of water = 0.800 L = 800 mL
since density of water 1 g/mL, its mass will be 800 g
m(solvent)= 800 g
= 0.8 Kg

Molality,
m = number of mol / mass of solvent in Kg
=(0.2698 mol)/(0.8 Kg)
= 0.337 molal
Answer: 0.337 molal