Question #de464

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Question #de464

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Answer 1 · 2015-01-04T21:02:19

The noble gas is $X e$ $Xe$ and its molar mass is $130.8$ $130.8$ $g/mol$ $"g/mol"$ .

What you need to know in order to solve this problem is that molar mass is defined as the mass of the gas divided by how much of the gas you have - the number of moles. So,

$M = \frac{m}{n} \Rightarrow n = \frac{m}{M}$ $M = m/n => n = m/M$

Starting from the ideal gas law, you can replace $n$ $n$ , the number of moles, with the above equation to get

$P V = \frac{m}{M} \cdot R T \Rightarrow M = m \cdot \frac{R T}{P V}$ $PV = m/M * RT => M = m * (RT)/(PV)$

If you look closer at this, you will notice that you have the formula for density in there as well. SInce

$ρ = \frac{m}{V}$ $rho = m/V$ , you get that

$M = \frac{m}{V} \cdot \frac{R T}{P} = ρ \cdot \frac{R T}{P}$ $M = m/V * (RT)/P = rho * (RT)/P$

At STP, the temperature is equal to 273.15 K, and the pressure is equal to 1.00 atm. Since the density of the gas is given to you, you'll get

$M = 5.84 \frac{g}{d m^{3}} \cdot \frac{0.082 \frac{L \cdot a t m}{K \cdot m o l} \cdot 273.15 K}{1.00 a t m}$ $M = 5.84 g/(dm^3) * (0.082 (L*atm)/(K * mol) * 273.15K)/(1.00atm)$

$M = 130.8 \frac{g}{m o l}$ $M = 130.8 g/(mol)$

(remember that $1 d m^{3} = 1 L$ $1dm^3 = 1L$ ).

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