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Question: What is the molar mass of a gas that has a density of \(1.02g{L^{ - 1}}\) at \(0.990\) atm pressure ...

What is the molar mass of a gas that has a density of 1.02gL−11.02g{L^{ - 1}} at 0.9900.990 atm pressure and 37∘C{37^ \circ }C ..

Explanation

Solution

Hint : To solve this question ,, we must know the ideal gas law :: PV=nRTPV = nRT .. Since the number of moles nn is not provided directly in the question ,, we must solve it using density ρ\rho ..Density is mass of unit volume of a substance that is mass divided by volume and number of moles is given mass divided by molar mass of the gas .. The value of universal gas constant will be 0.0821atmLmol−1K−10.0821atmLmo{l^{ - 1}}{K^{ - 1}} ..

Complete Step By Step Answer:
We know the ideal gas equation ,, PV=nRTPV = nRT ,, where P is the pressure of the gas ,, V is the volume of the gas ,, n is the number of moles of the gas ,, R is the universal gas constant and T is the temperature of the gas in Kelvin ..
Given ,, P == 0.9900.990atm
T == 37∘{37^ \circ }C == 37+273.15=310.15K37 + 273.15 = 310.15K
R == 0.0821atmLmol−1K−10.0821atmLmo{l^{ - 1}}{K^{ - 1}}
Since, volume and number of moles of the gas is not given ,, we have to find it out with the help of density equation ..
Number of moles ,, n=givenmass(m)Molarmass(M)n = \dfrac{{givenmass(m)}}{{Molarmass(M)}}
Density ,, ρ=mass(m)Volume(V)\rho = \dfrac{{mass(m)}}{{Volume(V)}}
Rewriting the ideal gas equation, PV=mMRTPV = \dfrac{m}{M}RT
M=mVRTPM = \dfrac{m}{V}\dfrac{{RT}}{P}
M=ρRTPM = \rho \dfrac{{RT}}{P}
Now, substituting the given values, we get the molar mass of the gas as ::
M=1.02gL−1×0.0821atmLmol−1K−1×310.15K0.990atmM = \dfrac{{1.02g{L^{ - 1}} \times 0.0821atmLmo{l^{ - 1}}{K^{ - 1}} \times 310.15K}}{{0.990atm}} == 26.23gmol−126.23gmo{l^{ - 1}}
Therefore ,, the molar mass of the gas is 26.23gmol−126.23gmo{l^{ - 1}} ..

Note :
The values of universal gas constant can vary according to the units of pressure ,, volume given in the question .. Here ,, since the pressure was in atm and volume in L ,,we substituted R value as 0.0821atmLmol−1K−10.0821atmLmo{l^{ - 1}}{K^{ - 1}} .. For e g ,, if the pressure was given in Joules and volume in L ,, R value will be 8.314Jmol−1K−18.314Jmo{l^{ - 1}}{K^{ - 1}} .. Also, while substituting temperature in the ideal gas law equation ,, it should be converted to Kelvin (K)(K) ..