Chapter 13, Problem 45A

Interpretation Introduction

Interpretation: -Volume of ${\mathrm{Cl}}_2(g)$ needs to be measured at STP which is needed to produce 10.0g of ${\mathrm{KMnO}}_4$ .

Concept Introduction: - For calculating the volume ofgas,first we calculate the number of moles of gas(n). Then volume is calculated using Ideal gas equation

  $\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

Firstly, we calculate the no. of moles and then putting values in the ideal gas equation for calculating volume.

Answer to Problem 45A

Volume of ${\mathrm{Cl}}_2(g)$ is $1.412L$

Explanation of Solution

There is a relationship between volume of a gas and number of molecules which expressed in Avogadro’s law

  $\mathrm{V}\propto N;n=\frac{N}{N_0}$

{here No = Avogadro’snumber}

  $\mathrm{n}\propto \mathrm{N}$

Volume of given mass of gas (v) is proportional to the number of moles(n)

  $\mathrm{N}\propto n$

  $$

Volume of $C{l}_2(g)$

  $\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

V =?

No. of moles of chlorine (n) $=\frac{\text{mass}}{\text{molar mass}}$

  $\mathrm{n}=\frac{10.0g}{158.03g/mol}=0.063moles.$

  $\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$ {at standard temperature and pressure}

  $\begin{array}{l}\text{V = }\frac{\text{0}\text{.063 mol × 0}\text{.0821atm L × 273k}}{\text{1atm × k mol}}\\ \text{V = 0}\text{.063 × 0}\text{.0821L × 273}\\ \text{V = 1}\text{.412 L}\end{array}$

Volume of ${\mathrm{Cl}}_2(g)$ is $1.412L$ .

Conclusion

Volume of ${\mathrm{Cl}}_2(g)$ is $1.412L$ .