A solution of glucose in water is labeled as 10% w/w. What should be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 gmL-1, then what shall be the molarity of the solution?

By BYJU'S Exam Prep

Updated on: September 25th, 2023

The molality and mole fraction of each component in the solution is 0.62M and 0.989. If the density of the solution is 1.2 gmL-1, then the molarity of the solution is 0.67 M. Steps to calculate the molarity of the solution:

Step 1:

10% w/w glucose solution denotes the presence of 10 g of glucose in a 100 g solution that also contains 90 g of water. The amount of moles of glucose in 1 kilogram of water is known as molality.

Molar mass of glucose (C6H12O6) = 6 x 12 + 12 x 1 + 6 x 16 = 180 gmol-1

We know that:

Number of moles = given mass in grams/ molecular mass

Number of moles of glucose = 10/180 mol = 0.056 mol

Again we have:

The molality of solution = number of moles/ weight of solvent

Substituting the values we get:

= 0.056mol/ 0.09kg

On simplifying we get:

= 0.62M

Number of moles of water = 90g/ 18 gmol-1 = 5 mol

Table of content

1. A solution of glucose in water is labeled as 10% w/w. What should be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 gmL-1, then what shall be the molarity of the solution?

Step 2: calculation of mole fraction

Mole fraction = number of moles of solute/ number of moles of solute + number of moles of solvent

Substituting the values we get:

Mole fraction of glucose, xg = 0.056/ (0.056 + 5) = 0.011

Again by substituting the values we get

Mole fraction of water, xw = 1 – xg = 1 – 0.011 = 0.989

Step 3:

If the density of the solution is 1.20 g/mL, then the volume of the 100 g solution can be given as:

100g/ 1.2 gmL-1 = 83.33 mL = 83.33 x 10-3 L

Substituting the values we get:

Molarity of the solution = 0.056 mol/ 83.33 x 10-3 L = 0.67 M

Summary: