Chapter 16, Problem 95A

Interpretation Introduction

Interpretation:

The mass of iron (III) chloride needed to prepare 1.00 L of a 0.22 M solution needs to be determined.

Concept introduction:

The molarity of a solution is given by the formula,

$\text{Molarity}\left(\text{M}\right)\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{litres}\text{of}\text{solution}}$

Answer to Problem 95A

The mass of iron (III) chloride needed to prepare 1.00 L of a 0.22 M solution is 14.2 g.

Explanation of Solution

Given, volume of solution = 1.00 L

Molarity=0.22 M

We have to first calculate the value of moles of solute in order to calculate the mass of iron (III) chloride using the formula, ‘

$\text{Molarity}\left(\text{M}\right)\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{litres}\text{of}\text{solution}}$

Using the given value of molarity and volume in the above formula we calculate the mass of iron (III) chloride.

So,

$\begin{array}{c}\text{Molarity}\left(\text{M}\right)\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{litres}\text{of}\text{solution}}\\ \text{0}\text{.255}\text{M}\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{1}\text{.00}\text{L}}\\ \left(\text{0}\text{.255}\right)\left(\text{1}\text{.00}\right)\text{=}\text{moles}\text{of}\text{solute}\\ \text{0}\text{.255}\text{=}\text{moles}\text{of}\text{solute}\end{array}$

So, required number of moles of iron (III) chloride is 0.255 mol.

So, we multiply 0.255 mol by the molar mass of $\text{55}\text{.8}\frac{\text{g}}{\text{mol}}$ to determine the mass of iron (III) chloride.

Therefore,

$\begin{array}{c}\text{Mass}\text{of}\text{iron=}\left(\text{0}\text{.255}\text{mol}\right)\left(\text{55}\text{.8}\frac{\text{g}}{\text{mol}}\right)\\ =\left(\text{0}\text{.255}\times \text{55}\text{.8}\right)\text{g}\\ \text{=}\text{14}\text{.2}\text{g}\end{array}$

Conclusion

The required mass of iron (III) chloride is 14.2 g.