What is the atomic mass of a gas that has $ρ = 2.50 \cdot g \cdot L^{- 1}$ $rho=2.50gL^-1$ at a pressure of $0.974 \cdot a t m$ $0.974atm$ , and a temperature of $371 \cdot K$ $371K$ ?

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Answer 1 · 2017-01-24T07:29:17

Given $P V = n R T$ $PV=nRT$ .......... $molar mass$ $"molar mass"$ $≅$ $~=$ $80 \cdot g \cdot m o l^{- 1}$ $80*g*mol^-1$

$\frac{P}{R T} = \frac{n}{V} = \frac{\frac{mass}{molar mass}}{V}$ $P/(RT)=n/V=("mass"/"molar mass")/V$

And thus $molar mass = \frac{mass of gas}{P V} \times R T$ $"molar mass"="mass of gas"/(PV)xxRT$

All I have done is manipulate the equation,

So $molar mass = \frac{mass}{V} \times \frac{R T}{P}$ $"molar mass"="mass"/(V)xx(RT)/P$

But $\frac{mass}{V} = density, ρ$ $"mass"/V="density", rho$

Finally, $molar mass = ρ \times \frac{R T}{P}$ $"molar mass"=rhoxx(RT)/P$

$(2.50*g*cancel(L^-1)xx0.0821*cancel(L)*cancel(atm)*cancel(K^-1)*mol^-1xx371*cancel(K))/(0.974*cancel(atm))$

$=??*g*mol^-1$

What is the atomic mass of a gas that has rho=2.50*g*L^-1ρ=2.50⋅g⋅L−1rho=2.50*g*L^-1 at a pressure of 0.974*atm0.974⋅atm0.974*atm, and a temperature of 371*K371⋅K371*K?