Chapter 4.1, Problem 4.4AFP

Interpretation Introduction

Interpretation:

The volume of $3.30M$ sucrose solution that contains $135\text{}\text{g}$ of solute is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the volume of the solution when the amount of compound in moles and molarity of solution are given is as follows:

$\text{Volume of solution}\left(\text{L}\right)=\text{moles of solute}\left(\text{mol}\right)\left(\frac{\text{1}\text{L of solution}}{\text{molarity of solution}\left(\text{mol}\right)}\right)$

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

$\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$