Chapter 5, Problem 5.71P

(a)

Interpretation Introduction

Interpretation:

The molar mass of each liquid should be calculated.

Concept Introduction :

Ideal gas law is represented as follows:

  $\text{PV = nRT}$

Here,

P − pressure

V − volume

n − number of moles

R − universal gas constant

T − temperature

Answer to Problem 5.71P

Liquid I = $62.5\text{ g/mol}$

Liquid II = $52.6\text{ g/mol}$

Liquid III = $66.7\text{ g/mol}$

Explanation of Solution

Sample I;

  $\begin{array}{l}\text{PV = nRT}\\ \text{0}\text{.05951 atm }\times \text{ 0}\text{.75 L = n }\times \text{ 0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 343}\text{.15 K}\\ \text{n = }\frac{\text{0}\text{.05951 atm }\times \text{ 0}\text{.75 L}}{\text{0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 343}\text{.15 K}}\\ \text{n = 0}\text{.0016 mol}\end{array}$

  $\text{M = }\frac{\text{m}}{\text{n}}=\frac{0.1000\text{ g}}{0.0016\text{ mol}}=62.5\text{ g/mol}$

Sample II;

  $\begin{array}{l}\text{PV = nRT}\\ \text{0}\text{.07045 atm }\times \text{ 0}\text{.75 L = n }\times \text{ 0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 343}\text{.15 K}\\ \text{n = }\frac{\text{0}\text{.07045 atm }\times \text{ 0}\text{.75 L}}{\text{0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 343}\text{.15 K}}\\ \text{n = 0}\text{.0019 mol}\end{array}$

  $\text{M = }\frac{\text{m}}{\text{n}}=\frac{0.1000\text{ g}}{0.0019\text{ mol}}=52.6\text{ g/mol}$

Sample III;

  $\begin{array}{l}\text{PV = nRT}\\ \text{0}\text{.05767 atm }\times \text{ 0}\text{.75 L = n }\times \text{ 0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 343}\text{.15 K}\\ \text{n = }\frac{\text{0}\text{.05767 atm }\times \text{ 0}\text{.75 L}}{\text{0}\text{.08206 L}\text{.atm/mol}\text{.K }\times \text{ 343}\text{.15 K}}\\ \text{n = 0}\text{.0015 mol}\end{array}$

  $\text{M = }\frac{\text{m}}{\text{n}}=\frac{0.1000\text{ g}}{0.0015\text{ mol}}=66.7\text{ g/mol}$

(b)

Interpretation Introduction

Interpretation:

The molecular formula of each sample should be determined.

Concept Introduction :

Molecular formula is a way of representing the number of moles of each atom present in a molecule.

Answer to Problem 5.71P

Sample I = ${\text{B}}_5{\text{H}}_9$

Sample II = ${\text{B}}_4{\text{H}}_{10}$

Sample III = ${\text{B}}_5{\text{H}}_{11}$

Explanation of Solution

Sample I; ${\text{B}}_x{\text{H}}_y$

Mass of boron in 1 mol of sample I = $62.5\text{ g}\times \frac{85.63}{100}=53.52\text{ g}$

  $\begin{array}{l}\frac{53.52\text{ g}}{10.81\text{ g/mol}}=x\\ x=4.95\simeq 5\end{array}$

Mass of hydrogen in 1 mol of sample I = $62.5\text{ g }\times \frac{100-85.63}{100}=8.98\text{ g}$

  $\begin{array}{l}\frac{8.98\text{ g}}{1.008\text{ g/mol}}=y\\ y=8.91\simeq 9\end{array}$

Therefore, molecular formula of sample I = ${\text{B}}_5{\text{H}}_9$

Sample II;

Mass of boron in 1 mol of sample II = $\text{52}\text{.6 g}\times \frac{81.10}{100}=42.66\text{ g}$

  $\begin{array}{l}\frac{\text{42}\text{.66 g}}{10.81\text{ g/mol}}=x\\ x=3.94\simeq 4\end{array}$

Mass of hydrogen in 1 mol of sample II = $\text{52}\text{.6 g }\times \frac{100-81.10}{100}=9.94\text{ g}$

  $\begin{array}{l}\frac{\text{9}\text{.94 g}}{1.008\text{ g/mol}}=y\\ y=9.86\simeq 10\end{array}$

Therefore, molecular formula of sample II = ${\text{B}}_4{\text{H}}_{10}$

Sample III;

Mass of boron in 1 mol of sample III = $\text{66}\text{.7 g}\times \frac{82.98}{100}=55.35\text{ g}$

  $\begin{array}{l}\frac{\text{55}\text{.35 g}}{10.81\text{ g/mol}}=x\\ x=5.12\simeq 5\end{array}$

Mass of hydrogen in 1 mol of sample III = $\text{66}\text{.7 g }\times \frac{100-82.98}{100}=11.35\text{ g}$

  $\begin{array}{l}\frac{\text{11}\text{.35 g}}{1.008\text{ g/mol}}=y\\ y=11.26\simeq 11\end{array}$

Therefore, molecular formula of sample III = ${\text{B}}_5{\text{H}}_{11}$

(c)

Interpretation Introduction

Interpretation:

The molecular formula of sample IV should be determined.

Concept Introduction :

The rate of effusion is related to molar mass as follows:

  $\begin{array}{l}\text{Rate of effusion }\propto \sqrt{\frac{1}{M}}\\ \\ M\text{ - molecular weight}\end{array}$

Answer to Problem 5.71P

  ${\text{B}}_2{\text{H}}_6$

Explanation of Solution

Rate of effusion of SO₂ = $\frac{250\text{ mL}}{13.04\text{ min}}=19.17\text{ mL/min}$

Rate of effusion of B_xH_y = $\frac{350\text{ mL}}{\text{12 min}}=29.17\text{ mL/min}$

  $\begin{array}{l}\frac{rate({\text{SO}}_2)}{rate({\text{B}}_x{\text{H}}_y\text{)}}=\sqrt{\frac{M_{B_x{H}_y}}{M_{S{O}_2}}}\\ \frac{19.17}{29.17}=\sqrt{\frac{M_{B_x{H}_y}}{64.06\text{ g/mol}}}\\ \sqrt{M_{B_x{H}_y}}=5.26\\ M_{B_x{H}_y}=27.67\text{ g/mol}\end{array}$

  $x=\frac{27.67\text{ g }\times \frac{78.14}{100}}{10.81\text{ g/mol}}=1.99\simeq 2$

  $y=\frac{27.67\text{ g }\times \frac{100-78.14}{100}}{1.008\text{ g/mol}}=6.00$

Therefore, formula of sample IV = ${\text{B}}_2{\text{H}}_6$