Chapter 6, Problem 5MCP

(a)

Interpretation Introduction

Interpretation:

$\%\left(m/V\right)$ of $5gMgC{l}_2$ in $2L$ solution has to be calculated.

Concept introduction:

$\%\left(m/V\right)$ can be calculated using the following equation.

    $\%\left(m/V\right)=\frac{massofsolute\left(g\right)}{volumeofsolution\left(mL\right)}\times 100\%$

Answer to Problem 5MCP

$\%\left(m/V\right)$ of $5gMgC{l}_2$ in $2L\left(2000mL\right)$ solution is $0.25\%.$

Explanation of Solution

$\%\left(m/V\right)$ of $5gMgC{l}_2$ in $2L\left(2000mL\right)$ solution can be calculated as follows,

    $\begin{array}{c}\%\left(m/V\right)=\frac{massofsolute\left(g\right)}{volumeofsolution\left(mL\right)}\times 100\%\\ \\ =\frac{5g}{2000mL}\times 100\%\\ \\ =0.25\%.\end{array}$

Percent mass/volume is $0.25\%.$

(b)

Interpretation Introduction

Interpretation:

$\%\left(m/m\right)$ of $5gMgC{l}_2$ in $2L$ solution has to be calculated.

Concept introduction:

$\%\left(m/m\right)$ can be calculated using the following equation.

    $\%\left(m/m\right)=\frac{massofsolute\left(g\right)}{massofsolution\left(g\right)}\times 100\%$

Answer to Problem 5MCP

$\%\left(m/V\right)$ of $5gMgC{l}_2$ in $1.02g/L$ solution is $245\%.$

Explanation of Solution

Given that, the volume and density of solution are $2L$ and $1.02g/L$ respectively.

So, mass of solution can be calculated as follows,

    $\begin{array}{c}massofsolution=volume\times density\\ \\ =2L\times \left(1.02g/L\right)\\ \\ =2.04g\end{array}$

$\%\left(m/m\right)$ of $5gMgC{l}_2$ in $2.04g$ solution can be calculated as follows,

    $\begin{array}{c}\%\left(m/m\right)=\frac{massofsolute\left(g\right)}{massofsolution\left(g\right)}\times 100\%\\ \\ =\frac{5g}{2.04g}\times 100\%\\ \\ =245\%.\end{array}$

Percent mass/mass is $245\%.$

(c)

Interpretation Introduction

Interpretation:

Molarity of $5gMgC{l}_2$ in $2L$ solution has to be calculated.

Concept introduction:

Molarity can be calculated as follows,

    $Molarity=\frac{moleofsolute}{Volumeofsolution\left(L\right)}$

Answer to Problem 5MCP

Molarity of $5gMgC{l}_2$ in $2L$ solution is $0.025M.$

Explanation of Solution

Number of moles of magnesium chloride can be calculated as follows,

  $\begin{array}{c}NumberofmolesofMgC{l}_2=\frac{givenmass}{molarmass}\\ \\ =\frac{5g}{95.211g/mole}\\ \\ =0.05mole.\end{array}$

Therefore molarity of magnesium chloride solution is given below.

    $\begin{array}{c}Molarity=\frac{moleofsolute}{Volumeofsolution\left(L\right)}\\ \\ =\frac{0.05mole}{2L}\\ \\ =0.025M.\end{array}$

Molarity is $0.025M.$

(d)

Interpretation Introduction

Interpretation:

Molarity of magnesium ion $5gMgC{l}_2$ in $2L$ solution has to be calculated.

Answer to Problem 5MCP

Molarity of magnesium ion $5gMgC{l}_2$ in $2L$ solution is $0.025M.$

Explanation of Solution

Number of moles of magnesium chloride can be calculated as follows,

  $\begin{array}{c}NumberofmolesofMgC{l}_2=\frac{givenmass}{molarmass}\\ \\ =\frac{5g}{95.211g/mole}\\ \\ =0.05mole.\end{array}$

$\text{1}\text{mole}$ Magnesium chloride contains $1mole$ magnesium ions.  So $0.05mole$ mole of magnesium chloride contains $0.05mole$ magnesium ions. 

Therefore molarity of magnesium ion in solution is given below.

    $\begin{array}{c}Molarity=\frac{moleofsolute}{Volumeofsolution\left(L\right)}\\ \\ =\frac{0.05mole}{2L}\\ \\ =0.025M.\end{array}$

Molarity of magnesium ion is $0.025M.$

(e)

Interpretation Introduction

Interpretation:

Molarity of chloride ion $5gMgC{l}_2$ in $2L$ solution has to be calculated.

Answer to Problem 5MCP

Molarity of chloride ion $5gMgC{l}_2$ in $2L$ solution is $0.05M.$

Explanation of Solution

Number of moles of magnesium chloride can be calculated as follows,

  $\begin{array}{c}NumberofmolesofMgC{l}_2=\frac{givenmass}{molarmass}\\ \\ =\frac{5g}{95.211g/mole}\\ \\ =0.05mole.\end{array}$

$\text{1}\text{mole}$ Magnesium chloride contains $2mole$ chloride ions.  So $0.05mole$ mole of magnesium chloride contains $0.1mole$ chloride ions. 

Therefore molarity of chloride ion in solution is given below.

    $\begin{array}{c}Molarity=\frac{moleofsolute}{Volumeofsolution\left(L\right)}\\ \\ =\frac{0.1mole}{2L}\\ \\ =0.05M.\end{array}$

Molarity of chloride ion is $0.05M.$

(f)

Interpretation Introduction

Interpretation:

Osmotic pressure of $5gMgC{l}_2$ in $2L$ solution has to be calculated.

Concept introduction:

Osmotic pressure ($\pi$) can be calculated as follows,

  $\pi =iMRT$

Where,

    $i$ is number of particles per mole of solute

    $M$ is molar concentration of the solute

    $R$ is universal gas constant

    $T$ is temperature in Kelvin scale.

Answer to Problem 5MCP

Osmotic pressure of $5gMgC{l}_2$ in $2L$ solution is $1.8atm.$

Explanation of Solution

Molarity of magnesium chloride solution is $\text{0}\text{.025}\text{M}\text{.}$

Given temperature is ${25}^{o}C\left(298.15K\right).$

The value of universal gas constant is $0.0821L.atm/K.mole.$

There will be three ions per mole of the solution, because magnesium chloride dissociates to give one magnesium ion and two chloride ions.

Therefore osmotic pressure can be calculated as follows,

    $\begin{array}{c}\pi =iMRT\\ \\ =\left(3\right)\times \left(\text{0}\text{.025}\text{M}\right)\times \left(0.0821L.atm/K.mole.\right)\times \left(298.15K\right)\\ \\ =1.8atm.\end{array}$

Osmotic pressure is $1.8atm.$