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The density of a gas is 0.259 g L1 at 400 K and 190 torr. Find its molar mass.

Answer
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Hint: We can solve this question by applying the ideal gas equation. The ideal gas equation gives the relationship between pressure of the gas, volume of the gas and temperature of the gas. Then using the equation of density, substitute the equation of volume in the ideal gas equation and solve for the molar mass.

Formulae Used:
1) PV=nRT
2) d=mV

Complete step-by-step solution:
We know the ideal gas equation is as follows:
PV=nRT …… (1)
Where P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of gas,
R is the universal gas constant,
T is the temperature.

We know that the density is the ratio of the molar mass to the volume. Thus,
d=mV
Where d is the density of the gas,
m is the molar mass of the gas,
V is the volume of the gas.

Rearrange the equation for the volume of the gas. Thus,
V=md
Thus, equation (1) becomes as follows:
P×md=nRT
P=dm×nRT
Rearrange the equation for the molar mass of the gas. Thus,
m=dP×nRT …… (2)
Use equation (2) to calculate the molar mass of the gas.
Convert the units of pressure from torr to atm using the relation as follows:
1 atm=760 torr
Thus,
P=190 torr×1 atm760 torr
P=0.25 atm
Thus, the pressure is 0.25 atm.

Calculate the molar mass of the gas using the equation (2) as follows:
In the equation (2), substitute 0.259 g L1 for the density of the gas, 0.25 atm for the pressure of the gas, 0.0821 L atm mol1 K1 or the universal gas constant, 400 K for the temperature of the gas. As the number of moles are not given, substitute 1 mol for the number of moles of gas and solve for the molar mass of the gas. Thus,
m=0.259 g L10.25 atm×1 mol×0.0821 L atm mol1 K1×400 K
m=34.02 g mol1

Thus, the molar mass of the gas is 34.02 g mol1.

Note: The units of pressure should be converted carefully from torr to atm. Remember that 1 atm=760 torr. Also, use proper units of the universal constant of gas to get the correct answers. The units of universal gas constant should be L atm mol1 K1.