Chapter 13, Problem 137QRT

(a)

Interpretation Introduction

Interpretation:

The concentration of ${\text{SnF}}_2$ in $\text{ppm}$ and $\text{ppb}$ has to be calculated.

Answer to Problem 137QRT

The concentration of ${\text{SnF}}_2$ in ppm and ppb are $6300ppm$ and $6,300,000ppb$, respectively.

Explanation of Solution

Given information is given as, $\text{0}\text{.63\%}\text{of}{\text{SnF}}_{\text{2}}$ is present.

As known, $\text{1\%}\text{=}\text{10,000 ppm}\text{.}$

So, $\text{0}\text{.63\%}\text{of}{\text{SnF}}_{\text{2}}$ becomes,

    $\text{0}\text{.63\%}{\text{SnF}}_{\text{2}}\times \frac{\text{10,000}\text{ppm}}{\text{1\%}}\text{=}\text{6300 ppm}{\text{SnF}}_{\text{2}}$.

Conversion of ppm to ppb as follows,

  $\text{6300 ppm}{\text{SnF}}_{\text{2}}\times \frac{\text{1000}\text{ppb}{\text{SnF}}_{\text{2}}}{\text{1ppm}{\text{SnF}}_{\text{2}}}\text{=}\text{6,300,000 ppb}{\text{SnF}}_{\text{2}}$.

Hence, the concentration of ${\text{SnF}}_2$ in ppm and ppb are $6300ppm$ and $6,300,000ppb$, respectively.

(b)

Interpretation Introduction

Interpretation:

The molarity of ${\text{SnF}}_2$ in solution has to be calculated.

Answer to Problem 137QRT

The molarity of ${\text{SnF}}_2$ in solution is $0.040M.$

Explanation of Solution

A sample of exactly $\text{100}\text{g}$ of solution contains $\text{0}\text{.63}\text{g}{\text{SnF}}_2$ and $\text{99}\text{.37}\text{g}$ water. So,

  $\frac{\text{0}\text{.63}\text{g}{\text{SnF}}_{\text{2}}}{{\text{10}}^{\text{2}}\text{g}\text{solution}}\text{×}\frac{\text{0}\text{.998}\text{g}\text{solution}}{\text{1}\text{mL}}\text{×}\frac{\text{1}\text{mol}{\text{SnF}}_{\text{2}}}{\text{156}\text{.707}\text{g}{\text{SnF}}_{\text{2}}}\text{×}\frac{\text{1000}\text{mL}}{\text{1}\text{L}}\text{=}\text{0}\text{.040}\text{M}{\text{SnF}}_{\text{2}}.$

Hence, molarity of ${\text{SnF}}_2$ in solution is $0.040M.$

(c)

Interpretation Introduction

Interpretation:

From one metric ton of cassiterite, number of $\text{250}\text{mL}$ bottles of $\text{0}\text{.63}\text{\%}{\text{SnF}}_2$ solution that can be prepared has to be calculated.

Answer to Problem 137QRT

The number of bottles calculated as $4.99\times 10^5.$

Explanation of Solution

The given reaction;

  ${\text{SnO}}_{\text{2}}\left(\text{s}\right)\text{+}\text{2C}\left(\text{s}\right)\to \text{Sn}\left(\text{s}\right)\text{+}\text{2CO}\left(\text{g}\right)$.

In exactly one metric ton, ${10}^6\text{g}$ is present; then theoretical value of Tin is,

  ${10}^6\text{g}{\text{SnO}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}\text{of}{\text{SnO}}_{\text{2}}}{\text{150}\text{.709}\text{g}{\text{SnO}}_{\text{2}}}\text{×}\frac{\text{1}\text{mol}\text{Sn }}{\text{1}\text{mol}{\text{SnO}}_{\text{2}}}\text{×}\frac{\text{118}\text{.710}\text{g}\text{Sn}}{\text{1}\text{mol}\text{Sn}}\text{=}\text{7}\text{.88}\times {\text{10}}^5\text{g}\text{Sn}\text{.}$

So, the actual mass of Tin is,

  $\text{7}{\text{.88×10}}^{\text{5}}\text{g}\text{Sn}\text{theoretical}\text{×}\frac{\text{80}\text{g}\text{Sn}\text{actual}}{\text{100}\text{g}\text{Sn}\text{theoretical}}\text{=}\text{6}\text{.30×}{\text{10}}^{\text{5}}\text{g}\text{Sn}\text{.}$

The theoretical mass of ${\text{SnF}}_{\text{2}}$ is,

  $6.30\times {10}^6\text{g}\text{Sn}\text{×}\frac{\text{1}\text{mol}\text{of}\text{Sn }}{\text{118}\text{.710g}\text{Sn}}\text{×}\frac{\text{1}\text{mol}{\text{SnF}}_2}{\text{1}\text{mol}\text{Sn}}\text{×}\frac{\text{156}\text{.707 g}{\text{SnF}}_2}{\text{1}\text{mol}{\text{SnF}}_2}\text{=}\text{8}\text{.32}\times {\text{10}}^5\text{g}{\text{SnF}}_2\text{.}$

The actual mass of ${\text{SnF}}_{\text{2}}$ is,

  $\text{8}{\text{.32×10}}^{\text{5}}\text{g}{\text{SnF}}_{\text{2}}\text{theoretical}\text{×}\frac{{\text{94 g SnF}}_{\text{2}}\text{actual}}{\text{100g}{\text{SnF}}_{\text{2}}\text{theoretical}}\text{=}\text{7}\text{.82}\text{×}{\text{10}}^{\text{5}}\text{g}{\text{SnF}}_{\text{2}}\text{.}$

Number of bottles is calculated by using bottle volume; which is calculated by molarity obtained

  $\text{7}\text{.82}{\text{×10}}^{\text{5}}{\text{g SnF}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{SnF}}_{\text{2}}}{\text{156}\text{.707}\text{g}{\text{SnF}}_{\text{2}}}\text{×}\frac{\text{1L}}{\text{0}\text{.040}{\text{mol SnF}}_{\text{2}}}\text{×}\frac{\text{1000}\text{mL}}{\text{1}\text{L}}\text{×}\frac{\text{1Bottle}}{\text{250}\text{mL}}\text{=}4.99\times 10^{5}bottles.$

Hence, number of bottles calculated as is $4.99\times 10^5.$