What is the atomic mass of a gas that has $rho=2.50gL^-1$ at a pressure of $0.974atm$ , and a temperature of $371K$ ?

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Answer 1 · 2017-01-24T07:29:17

Given $PV=nRT$ .......... $"molar mass"$ $~=$ $80*g*mol^-1$

$P/(RT)=n/V=("mass"/"molar mass")/V$

And thus $"molar mass"="mass of gas"/(PV)xxRT$

All I have done is manipulate the equation,

So $"molar mass"="mass"/(V)xx(RT)/P$

But $"mass"/V="density", rho$

Finally, $"molar mass"=rhoxx(RT)/P$

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

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

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