Chapter 12, Problem 100A

Interpretation Introduction

Interpretation: The volume in milliliter required to make $1.5$ liters of $0.1$ molar $\mathrm{KBr}$ solution needs to be determined. Also, the boiling point of the new solution needs to be calculated.

Concept introduction: Molarity is calculated as the number of moles of solute in one liter of a solvent.

Answer to Problem 100A

The calculated volume to make $1.5$ liters of $0.1$ molar $\mathrm{KBr}$ solution is $300$ milliliters.

The boiling point of new solution is $100.1$ in degree Celsius.

Explanation of Solution

Given information: Solution containing $135.2$ gram of dissolved $\mathrm{KBr}$ in $2.3$ litre of water.

As molarity is the number of moles of solute in one liter of a solvent so by using this formula we calculated the molarity of the original solution and used this further to calculate the volume required to make $1.5$ liters of $0.1$ molar $\mathrm{KBr}$ solution.

Molarity of original solution= $135.2\times \frac{1}{119}=1.14\text{mol }\mathrm{KBr}$

$\text{M}=\frac{1.14}{2.3}=0.496\text{M}$

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Now, diluting the solution, the volume required, $\text{V}=0.10\times \frac{1.5}{0.496}=0.30\text{L}=300\text{mL}$ .

Now, the boiling point of the new solution, $\Delta T_b=im{K}_b$ .

Here, m is molality, i is Van’t Hoff factor and $K_b$ is molal elevation constant.

For KBr, Van’t Hoff factor will be 2.

For water, the value of $K_b$ is 0.512 ^oC/m.

The molality will be the same as molarity in this case because the volume of the solution is equal to the volume of water. Since the density of water is 1 g/mL, thus, the mass of water will be equal to that of volume.

Therefore, $\Delta T$ = $2\times 0.10\times 0.512=0.10°\text{C}$

Now, boiling point of new solution, $T_b$ = $100°\text{C}+0.10\text{°C}$ $=100.10\text{°C}$

Conclusion

The calculated volume to make $1.5$ litre of $0.1$ molar $\mathrm{KBr}$ solution is $300$ millilitres.

The boiling point of the new solution is $100.1$ in degree Celsius.