Chapter 15, Problem 44A

(a)

Interpretation Introduction

Interpretation:

The normality of the given solution is to be calculated.

Concept Introduction:

The normality of the solution is given as follows:

  $normality=\frac{numberofequivalentsofsolute}{Volumeofsolutioninlitre}$

Answer to Problem 44A

  $0.277\text{N}$

Explanation of Solution

Given information:

The volume is, $V=1\text{0}\text{.2 mL}$

The mass is, $m=0.11\text{3 g}$ of the $\text{NaOH}$ solution.

Calculation:

Normality can be calculated as follows:

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litre}}$

The equivalent weight of $\text{NaOH}$ solution = molar mass $\text{NaOH}$ = $40.0\text{0 g}$

Thus, number of equivalents of solute will be:

  $\text{0}\text{.113g}\text{NaOH×}\frac{\text{1}\text{equivalent}\text{NaOH}}{\text{40}\text{.00g}}\text{=2}{\text{.83×10}}^{\text{-3}}\text{equiv}\text{NaOH}$

Now use the volume and equivalent weight in the formula as follows:

  $\begin{array}{l}N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution in}\text{litre}}\\ \text{=}\frac{\text{2}{\text{.83×10}}^{\text{-3}}\text{equiv}}{\text{10}{\text{.2×10}}^{{}^{\text{-3}}}\text{L}}\\ =0.277\mathrm{N}\end{array}$

Thus, the normality is $0.277\text{N}$ .

(b)

Interpretation Introduction

Interpretation:

The normality of the given solution is to be calculated.

Concept Introduction:

The normality of the solution is given as follows:

  $normality=\frac{numberofequivalentsofsolute}{Volumeofsolutioninlitre}$

Answer to Problem 44A

  $3.37\times {10}^{-3}\text{N}$

Explanation of Solution

Given information:

The volume is, $V=100.0\text{mL}$

The mass is, $m=\text{12}\text{.5mg=12}{\text{.5×10}}^{\text{-3}}\text{g}$ of the $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution.

Calculation:

The normality can be calculated as follows:

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{Volume of}\text{solution}\text{in}\text{litre}}$

The equivalent weight of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ solution = $\frac{\text{molar}\text{mass}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{2}}\text{=}\frac{\text{74}\text{.10}}{\text{2}}\text{=37}\text{.05g}$

  $\left(\text{12}\text{.5}{\text{×10}}^{\text{-3}}\text{g}\right)\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\text{×}\frac{\text{1}\text{equivalent}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{37}\text{.05g}}\text{=3}{\text{.37×10}}^{\text{-4}}\text{equiv}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$

Now use the volume and equivalent weight in the formula is,

  $\begin{array}{l}N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{volume}\text{of}\text{solution}\text{in}\text{litre}}\\ \text{=}\frac{\text{3}{\text{.37×10}}^{\text{-4}}\text{equiv}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{100}{\text{.0×10}}^{{}^{\text{-3}}}\text{L}}\\ \text{=}\text{3}{\text{.37×10}}^{\text{-3}}\text{N}\end{array}$

Thus, the normality is $3.37\times {10}^{-3}\text{N}$ .

(c)

Interpretation Introduction

Interpretation:

The normality of the given solution is to be calculated.

Concept Introduction:

The normality of the solution is given as follows:

  $normality=\frac{numberofequivalentsofsolute}{Volumeofsolutioninlitre}$

Answer to Problem 44A

  $1.63\text{N}$

Explanation of Solution

Given information:

The volume is, $V=155.0\text{mL}$

The mass is, $m=12.4\text{g}$ of the ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution.

Calculation:

The normality is calculated as follows:

  $\text{normality}=N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{nuber}\text{of}\text{solution}\text{of}\text{litre}}$

The equivalent weight of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution = $\frac{\text{molar}\text{mass}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{2}}\text{=}\frac{\text{98}\text{.09}}{\text{2}}\text{=49}\text{.05g}$

  $\text{12}\text{.4g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\frac{\text{1}\text{equivalent}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{49}\text{.05g}}\text{=0}\text{.253}\text{equiv}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Now use the volume and equivalent weight in the formula as:

  $\begin{array}{l}N=\frac{\text{number}\text{of}\text{equivalents}\text{of}\text{solute}}{\text{nuber}\text{of}\text{solution}\text{of}\text{litre}}\\ \text{=}\frac{\text{0}\text{.253}\text{equiv}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{155}{\text{.0×10}}^{{}^{\text{-3}}}\text{L}}\\ =1.63{\rm N}\end{array}$

Thus, the normality is $\text{1}\text{.63N}$ .