Chapter 8, Problem 33Q

(a)

Interpretation Introduction

Interpretation:

A preparation method for two liters of $\text{1}\text{.50 M KOH}$ has to be explained.

Concept Introduction:

The number of moles of solute in a solution calculated as

  $\text{Mole}\text{=}\frac{\text{Given}\text{weight}}{\text{Molar}\text{mass}}$

Molarity can be defined as the moles of solute to the volume of the solution (in liters). The molarity of a solution can be given by the formula,

    $\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute (in}\text{mol)}}{\text{Volume}\text{of}\text{solution (in}\text{L)}}$

Explanation of Solution

The mass of solid potassium hydroxide required to prepare two litres of $\text{1}\text{.50M}$ potassium hydroxide is calculated as,

    $\begin{array}{l}\text{Moles}\text{=}\text{2L}\text{×}\frac{\text{1}\text{.50mol}}{\text{1L}}\text{=}\text{3}\text{.00mol}\\ \\ \text{mass}\text{of}\text{KOH}\text{=}\text{3}\text{.00mol}\text{×}\frac{\text{56}\text{.1056 g}}{\text{1mol}}\\ \\ \text{=}\text{168}\text{.32g}\end{array}$

Weigh out $\text{168}\text{.32g}$ of potassium hydroxide in a two litre volumetric flask.  Fill the flask with distilled water up to two litre mark.

(b)

Interpretation Introduction

Interpretation:

A preparation method for one liter of $\text{0}\text{.050 M NaBr}$ has to be explained.

Concept Introduction:

The number of moles of solute in a solution calculated as

  $\text{Mole}\text{=}\frac{\text{Given}\text{weight}}{\text{Molar}\text{mass}}$

Molarity can be defined as the moles of solute to the volume of the solution (in liters). The molarity of a solution can be given by the formula,

    $\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute (in}\text{mol)}}{\text{Volume}\text{of}\text{solution (in}\text{L)}}$

Explanation of Solution

The mass of solid sodium bromide required to prepare one litre of $\text{0}\text{.050 M NaBr}$ is calculated as,

    $\begin{array}{l}\text{Moles}\text{=}\text{1L}\text{×}\frac{\text{0}\text{.050mol}}{\text{1L}}\text{=}\text{0}\text{.050mol}\\ \\ \text{mass}\text{of}\text{NaBr}\text{=}\text{0}\text{.050mol}\text{×}\frac{\text{102}\text{.894 g}}{\text{1mol}}\\ \\ \text{=}\text{5}\text{.145g}\end{array}$

Weigh out $5.145\text{g}$ of sodium bromide in a one litre volumetric flask.  Fill the flask with distilled water up to one litre mark.

(c)

Interpretation Introduction

Interpretation:

A preparation method for $\text{0}\text{.10 L of 1}{\text{.2 M Mg(OH)}}_{\text{2}}$ has to be explained.

Concept Introduction:

The number of moles of solute in a solution calculated as

  $\text{Mole}\text{=}\frac{\text{Given}\text{weight}}{\text{Molar}\text{mass}}$

Molarity can be defined as the moles of solute to the volume of the solution (in liters). The molarity of a solution can be given by the formula,

    $\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute (in}\text{mol)}}{\text{Volume}\text{of}\text{solution (in}\text{L)}}$

Explanation of Solution

The mass of solid magnesium hydroxide required to prepare $\text{0}\text{.10 L of 1}{\text{.2 M Mg(OH)}}_{\text{2}}$ is calculated as,

    $\begin{array}{l}\text{Moles}\text{=}\text{0}\text{.10L}\text{×}\frac{\text{1}\text{.20mol}}{\text{1L}}\text{=}\text{0}\text{.120mol}\\ \\ \text{mass}\text{of}\text{Mg}{\left(\text{OH}\right)}_{\text{2}}\text{=}\text{0}\text{.120mol}\text{×}\frac{\text{58}\text{.3197 g}}{\text{1mol}}\\ \\ \text{=}\text{7}\text{.0g}\end{array}$

Weigh out $7.0\text{g}$ of magnesium hydroxide in a $\text{100mL}$ volumetric flask.  Fill the flask with distilled water up to $\text{100mL}$ mark.