Chapter 16, Problem 103A

Interpretation Introduction

Interpretation: The procedure of making $100\text{mL}$ of $0.50\text{M}$ $\text{KCl}$ starting with a stock solution $2.0\text{ M}$ $\text{KCl}$ is to be stated.

Concept Introduction: The molarity of a solution is obtained by dividing the moles of solute (the substance which is added to the solvent) by the volume of the solution and the solvent here is water.

Write the formula of volume for the standard solution.

  $V_s{C}_s=V_w{C}_w$   ...... (1)

Here,

  $V_s$ is Volume of stock solution

  $C_s$ is Concentration of stock solution

  $V_w$ is Volume of standard solution

  $C_w$ is Concentration of the standard solution

Answer to Problem 103A

To prepare $100\text{mL}$ of $0.50\text{ M}$

  $\text{KCl}$ , add $25\text{mL}$ of stock solution to water and make up the overall volume up to $100\text{mL}$ .

Explanation of Solution

Given:

The stock solution is $2.0\text{M}$ $\text{KCl}$ .

  $100\text{mL}$ of $0.50\text{M}$ $\text{KCl}$

Step 01: Calculate the volume for the standard solution.

Substitute, $2.0\text{M}$ for $C_s$ , $100\text{mL}$ for $V_w$ and $0.50\text{M}$ for $C_w$ in equation (1).

  $\begin{array}{c}{V}_s{C}_s=V_w{C}_w\\ V_s\times 2.0\text{M}=100\text{mL}\times 0.50\text{M}\\ V_s=\frac{100\text{mL}\times 0.50\text{M}}{2.0\text{M}}\\ =25\text{mL}\end{array}$

Hence, to prepare $100\text{mL}$ of $0.50\text{M}$ $\text{KCl}$ , add $25\text{mL}$ of stock solution to water and make up the overall volume up to $100\text{mL}$ .