Chapter 13, Problem 47QRT

(a)

Interpretation Introduction

Interpretation:

The molarity of the solute $\text{6}\text{.18g}{\text{MgNH}}_{\text{4}}{\text{PO}}_{\text{4}}\text{ in 250mL}$ solution has to be calculated.

Concept introduction:

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

  $\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}$

The number of moles of any substance can be determined using the equation.

  $\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Moles: One mole is equivalent to the mass of the substance consists of same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

Answer to Problem 47QRT

The molarity obtained as $0.18M.$

Explanation of Solution

Given mass is $6.18\text{g}$; Volume is $\text{250 mL}$; Molar mass ${\text{MgNH}}_{\text{4}}{\text{PO}}_{\text{4}}\text{is}\text{137}\text{.31 g/mol}.$

Calculate the number of moles:

  $\begin{array}{c}\text{Number of moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \\ \text{=}\frac{\text{6}\text{.18}\text{g}}{\text{137}\text{.31 g/mol}}\\ \\ \text{=}\text{45}\text{.00×}{\text{10}}^{\text{-3}}\text{moles}\text{.}\end{array}$

Calculate the molarity:

  $\begin{array}{c}\text{ Molarity}\text{ =}\frac{\text{Number of moles}}{\text{volume}}\\ \\ \text{=}\frac{\text{45}\text{.00×}{\text{10}}^{\text{-3}}\text{moles}}{\text{250}\text{ml}}\\ \\ \text{=}0.18M\text{.}\end{array}$

Therefore, the molarity of the solute $\text{6}\text{.18g}{\text{MgNH}}_{\text{4}}{\text{PO}}_{\text{4}}\text{ in 250mL}$ solution was $0.18M.$

(b)

Interpretation Introduction

Interpretation:

The molarity of the solute $\text{16}\text{.8g}{\text{NaCH}}_3\text{COO in 300mL}$ solution has to be calculated.

Concept introduction:

Refer part (a).

Answer to Problem 47QRT

The molarity obtained as $0.6826M\text{.}$

Explanation of Solution

Given mass is $\text{16}\text{.8}\text{g}$; Volume is $300\text{ mL}$; Molar mass ${\text{NaCH}}_3\text{COO}\text{is 82}\text{.0326 g/mol}\text{.}$

Calculate the number of moles:

  $\begin{array}{c}\text{Number of moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \\ \text{=}\frac{\text{16}\text{.8}\text{g}}{\text{82}\text{.0326 }\text{g/mol}}\\ \\ \text{=}\text{204}\text{.79×}{\text{10}}^{\text{-3}}\text{moles}\text{.}\end{array}$

Calculate the molarity:

  $\begin{array}{c}\text{ Molarity}\text{ =}\frac{\text{Number of moles}}{\text{volume}}\\ \\ \text{=}\frac{\text{204}\text{.79×}{\text{10}}^{\text{-3}}\text{moles}}{\text{300}\text{ml}}\\ \\ \text{=}0.6826M\text{.}\end{array}$

Therefore, the molarity of the solute $\text{16}\text{.8g}{\text{NaCH}}_3\text{COO in 300mL}$ solution was $0.6826M\text{.}$

(c)

Interpretation Introduction

Interpretation:

The molarity of the solute $\text{2}\text{.50g}{\text{CaC}}_2{\text{O}}_4\text{ in 750mL}$ solution has to be calculated.

Concept introduction:

Refer part (a).

Answer to Problem 47QRT

The molarity obtained as $0.026M.$

Explanation of Solution

Given mass is $\text{2}\text{.50}\text{g}$; Volume is $750\text{ mL}$; Molar mass ${\text{CaC}}_2{\text{O}}_4\text{is 128}\text{.097 g/mol}\text{.}$

Calculate the number of moles:

  $\begin{array}{c}\text{Number of moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \\ \text{=}\frac{\text{2}\text{.50}\text{g}}{\text{128}\text{.097 g/mol}}\\ \\ \text{=}\text{19}\text{.5×}{\text{10}}^{\text{-3}}\text{moles}\text{.}\end{array}$

Calculate the molarity:

  $\begin{array}{c}\text{ Molarity}\text{ =}\frac{\text{Number of moles}}{\text{volume}}\\ \\ \text{=}\frac{\text{19}\text{.5×}{\text{10}}^{\text{-3}}\text{moles}\text{.}}{\text{750}\text{ml}}\\ \\ \text{=}0.026M.\end{array}$

Therefore, the molarity of the solute $\text{16}\text{.8g}{\text{CaC}}_2{\text{O}}_4\text{ in 300mL}$ solution was $0.026M.$

(d)

Interpretation Introduction

Interpretation:

The molarity of the solute $\text{2}\text{.20g}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\text{ in 400mL}$ solution has to be calculated.

Concept introduction:

Refer part (a).

Answer to Problem 47QRT

The molarity obtained as $0.0415M.$

Explanation of Solution

Given mass is $\text{2}\text{.20}\text{g}$; Volume is $40\text{0 mL}$; Molar mass ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\text{is}\text{132}\text{.14 g/mol}\text{.}$

Calculate the number of moles:

  $\begin{array}{c}\text{Number of moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \\ \text{=}\frac{\text{2}\text{.20}\text{g}}{\text{132}\text{.14 g/mol}}\\ \\ \text{=}\text{16}\text{.6×}{\text{10}}^{\text{-3}}\text{moles}\text{.}\end{array}$

Calculate the molarity:

  $\begin{array}{c}\text{ Molarity}\text{ =}\frac{\text{Number of moles}}{\text{volume}}\\ \\ \text{=}\frac{\text{16}\text{.6×}{\text{10}}^{\text{-3}}\text{moles}}{\text{400}\text{ml}}\\ \\ \text{=}0.0415M.\end{array}$

Therefore, the molarity of the solute $\text{2}\text{.20g}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\text{ in 400mL}$ solution was $0.0415M.$