Question

A gas is found to have a density of 1.80 g/litre at 1 atm pressure and 27°c. the gas will be

Answer 1

Here is the correct answer

Answer 2

Answer : The gas will be Carbon dioxide.

Solution : Given,

Density of gas = 1.80 g/L

Pressure of gas = 1 atm

Temperature of gas = $27^oC=273+27=300K$      $(0^oC=273K)$

using ideal gas law,

$PV=nRT\\\\PV=\frac{w}{M}\times RT\\\\M=\frac{w}{V}\times \frac{RT}{P}\\\\M=\rho\times \frac{RT}{P}$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of the gas

R = gas constant = 0.0821 Latm/moleK

M = molar mass of gas

w = mass of gas

$\rho$ = density

Now put all the given values in the above formula, we get

$M=\rho\times \frac{RT}{P}$

$M=(1.80g/L)\times \frac{(0.0821Latm/moleK)\times (300K)}{1atm}=44.334g/mole$

Therefore, the molar mass of gas is equal to 44.334 grams and the gas will be carbon dioxide.