# A solution of glucose in water is labelled as 10% (W/W). - Chemistry

A solution of glucose in water is labelled as 10% (W/W).

Calculate:

a. Molality

b. Molarity of the solution.

[Given: Density of solution is 1.20 g mL-1 and molar mass of glucose is 180 g mol-1 ]

#### Solution

Given: Glucose solution is 10% W/W, Density of solution = 1.20 g mL-1

To find: a. Molality b. Molarity

1. "Molality" = "Number of moles of solute"/"Mass of solvent in kg"

2. "Molarity" = "Number of moles of solute"/"Volumeof solution in L"

Calculation: 10 % W/W of glucose solution means 100 g of solution contains 10 g of glucose.

Mass of glucose = 10 g
Mass of solution = 100 g
Mass of H2O = 100 – 10 = 90 g = 0.090 kg
Molar mass of glucose = 180 g mol-1

"Number of moles of glucose" = " Massof glucose"/"Molar massof glucose"

=(10 g)/(180 g mol^-1)=0.0556 mol

From formula (1),

"Molality of glucose solution "="Number of molesof glucose"/"Massof H2Oin kg"

= 0.0556 mol/0.090 kg
= 0.618 m

b. "Density of solution" =" Massof solution"/"Volumeof solution"

Volume of solution = 100g/1.20g mL-1 = 83.33 mL = 0.08333 L

"Molarity of glucose solution "= "Number of molesof glucose"/"Volume of solution in L"

= 0.0556 mol/0.08333 L
= 0.67 M

Concept: Colligative Properties and Determination of Molar Mass - Introduction
Is there an error in this question or solution?