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Question: The density of a gas is \[1.964gd{m^{ - 3}}\] at \[273K\] and \[76cmHg\]. The gas is: A.\[C{H_4}\]...

The density of a gas is 1.964gdm31.964gd{m^{ - 3}} at 273K273K and 76cmHg76cmHg. The gas is:
A.CH4C{H_4}
B.C2H6{C_2}{H_6}
C.CO2C{O_2}
D.XeXe

Explanation

Solution

We can solve the problem through the ideal gas situation. We should know what his ideal gas equation is. We will also calculate the molar mass of all compounds present in the above question. The ideal gas equation is only used for substances which are in a gaseous state.

Complete step by step solution: The density of a gas: We can define it as the mass per unit volume of a substance under specific conditions of temperature and pressure. In addition, the density of a gas is equal to its mass divided by its volume. Moreover, you can calculate the molar mass of a substance once you know the density of a gas. Also, the density of a gas varies with pressure and temperature.
In the question,
density of gas=1.964gdm31.964gd{m^{ - 3}}
Converting the above quantity in centimeters, we get
density of gas=1.964×103gcm31.964 \times {10^{ - 3}}gc{m^{ - 3}}
Temperature,T=273KT = 273K, pressure P=76cmHg=1atmP = 76cmHg = 1atm
Let the gas behaves ideally, so we know that for ideal gas;
PV=nRTPV = nRT; R=R = universal gas constant, V=V = volume
n=n = number of moles,
n=W(GivenMass)M(MolarMass)\Rightarrow n = \dfrac{{W(GivenMass)}}{{M(MolarMass)}}
PV=WMRTPV = \dfrac{W}{M}RT
Now interchange and ;
M=WVRTPM = \dfrac{W}{V}\dfrac{{RT}}{P}
Now we know that density, d=WPd = \dfrac{W}{P}
Applying it in the equation we get;
M=dRTPM = \dfrac{{dRT}}{P} -(1)
Now,
R=0.082LatmK1mol1R = 0.082Latm{K^{ - 1}}mo{l^{ - 1}}
1L=103cm31L = {10^3}c{m^3}
Putting all the above data in the equation(1) we get

M=1.964×103×0.082×103×273 M=1.964×0.082×273 M=43.97 44g  M = 1.964 \times {10^{ - 3}} \times 0.082 \times {10^3} \times 273 \\\ \Rightarrow M = 1.964 \times 0.082 \times 273 \\\ \Rightarrow M = 43.97 \\\ \approx 44g \\\

So the molar mass of the gas is 44g44g from the molar mass. We find out the gas, so let’s check the molar mass of all the gases which are present in the options.
The molar mass of CH4=12+(4×1)=30C{H_4} = 12 + (4 \times 1) = 30
The molar mass of C2H6=12×2+(6×1)=30{C_2}{H_6} = 12 \times 2 + (6 \times 1) = 30
The molar mass of CO2=12+(16×2)=44C{O_2} = 12 + (16 \times 2) = 44
So CO2C{O_2} has the molar mass of 44g44g.

The correct answer is option (C).

Note: The density of gases depends upon the temperature. The higher the temperature, the more the molecules are speed spread out and the lower the density as shown in the equation. The result is warm gases rise and cool gases sink. When pressure increases, density increases. When the pressure decreases, density decreases.