3.4 Number of Moles and Its Molar Volume

Number of Moles and Its Molar Volume
■ Definition : : Molar volume : The volume occupied by 1 mole of gas at a particular temperature and pressure.

	►	At room conditions (25°C and pressure 1 atmosphere), 1 mole of gas occupies a volume of 24dm3 .
	►	At standard condition (0 and pressure 1 atmosphere), 1 mole of gas occupies a volume 22.4dm3.

■ Summary of relationship between the number of moles, the volume of gas and the molar volume of gas.

✍ Worked-example 3.4(a)

	Calculate the volume occupied by 3.5 moles of ammonia gas, ${\text{NH}}_3$ at s.t.p. Solution
	Volume of gas $\mathrm{=\; 3.5}\times 22.4$ = 78.4dm3

✍ Worked-example 3.4(b)

	Calculate the volume occupied by the following gases at s.t.p: [RAM: H = 1, C = 12, N = 14, O = 16,], [NA = $6.02\times {10}^{23}$ , 1 mole gas occupies 22.4dm³ at s.t.p]
	(a)	8.8 g carbon dioxide, ${\text{CO}}_2$ Solution Step 1: $\begin{array}{c}\text{Number of moles =}\frac{\text{Mass of Carbon Dioxide}}{\text{Relative molecular mass}}\end{array}$ $\begin{array}{c}\text{=}\frac{8.8}{\mathrm{12\; +\; (2}\times \mathrm{16)}}\end{array}$ $\begin{array}{c}\text{=}\frac{8.8}{\mathrm{12\; +\; 32}}\end{array}$ $\begin{array}{c}\text{=}\frac{8.8}{44}\end{array}$ = 0.2mol Step 2: Volume of 8.8 g carbon dioxide, ${\text{CO}}_2$ $\text{= Number of mole}\times \text{Molar volume}$ $\mathrm{=\; 0.2}\times 22.4$ = 4.48dm3
	(b)	$1.5\times {10}^{23}$ molecules of oxygen, ${\text{O}}_2$ Solution Step 1: $\begin{array}{c}\text{Number of moles =}\frac{\text{Number of particles}}{\text{Avogrado constant}}\end{array}$ $\begin{array}{c}\text{=}\frac{1.5\times {10}^{23}}{6\times {10}^{23}}\end{array}$ = 0.25mol Step 2 Volume of oxygen $\text{= Number of mole}\times \text{Molar volume}$ $\mathrm{=\; 0.25}\times 22.4$ = 5.6dm3

Summary of Mole Concept
■ Summary of relationship between the number of moles of compound with the number of particles, particle mass and gas volume.

✍ Worked-example 3.4(c)

	Calculate the volume occupied by the following gases at s.t.p. [Molar volume at s.t.p = 22.4dm3, NA = $6\times {10}^{23}$ , H = 1, C = 12, O = 16, N = 14, Cl = 35.5, Ne = 20]
	(a)	0.5 g of neon , Ne. Solution Step 1: $\begin{array}{c}\text{Number of moles =}\frac{\text{Mass of neon}}{\text{Relative atom mass}}\end{array}$ $\begin{array}{c}\text{=}\frac{0.5}{20}\end{array}$ = 0.025mol Step 2: Volume of 0.5 g neon ${\text{CO}}_2$ $\text{= Number of mole}\times \text{Molar volume}$ $\mathrm{=\; 0.025}\times 22.4$ = 0.56dm3
	(b)	1.5 moles of carbon dioxide, ${\text{CO}}_2$ Solution : Volume of 1.5 moles of ${\text{CO}}_2$ $\text{= Number of mole}\times \text{Molar volume}$ $\mathrm{=\; 1.5}\times 22.4$ = 33.6dm3
	(c)	$4.8\times {10}^{22}$ molecules Chloride, ${\text{Cl}}_2$ Solution Step 1: $\begin{array}{c}\text{Number of moles =}\frac{\text{Number of particles}}{\text{Avogrado constant}}\end{array}$ $\begin{array}{c}\text{=}\frac{4.8\times {10}^{22}}{6\times {10}^{23}}\end{array}$ = 0.08mol Step 2: Volume of ${\text{Cl}}_2$ $\text{= Number of mole}\times \text{Molar volume}$ $\mathrm{=\; 0.08}\times 22.4$ = 1.792dm3

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