Chapter 16, Problem 16.4P

Interpretation Introduction

Interpretation:

Preparation of $500.0g$ of an aqueous solution of sucrose in which the mole fraction of sucrose is $0.125$ has to be described.

Concept Introduction:

Mole fraction:

Mole fraction of a solute in a solution can be calculated as given below.

  ${\chi }_1=\frac{n_1}{n_{tot}}$

Where, $n_1$ is the number of moles of solute and $n_{tot}$ is the sum of moles of solute and solvent.