Chapter 4, Problem 30PE

Interpretation Introduction

Interpretation:

The number of grams of sodium sulfate required to prepare given volume of $\text{0}\text{.0150 M }$ sodium sulfate solution is to be determined.

Concept Introduction:

The relation between liter (L) and milliliter (mL) is given as: 1L = 1000 mL.

The number of moles present in one liter of a solution is known as the molarity of the solution. It is calculated as follows:

$M=\frac{n}{V}$

Here, $M$ is the molarity of the solution, $V$ is the volume of the solution in liters and $n$ is the number of moles of the solute.

The number of moles of the solute is calculated by dividing the mass of the solute by the molar mass of the solute.

$n=\frac{m}{M}$

Here, $m$ is the mass of the solute and $M$ is the molar mass of the solute.