Asample of gas with a volume of 2.0 l at stp is found to have a mass of 1.43 g calculate the molecular mass of this gas. is it possible this gas is methane? why or why not

molar mass = 16.03 g/mol.

Yes, the mentioned gas is methane.

Explanation:

where, P is the pressure of the gas in atm (P = 1.0 atm, at STP).

V is the volume of the gas in L (V = 2.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R  is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 273 K, at STP).

∴ n = PV/RT = (1.0 atm)(2.0 L)/(0.0821 L.atm/mol.K)(273 K) = 0.08923 mol.

∵ n = mass/molar mass

∴ molar mass = mass/n = (1.43 g)/(0.08923 mol) = 16.03 g/mol.

∵ molar mass of methane (CH₄) = 16.0 g/mol.

So, the mentioned gas is methane.

a:

by the combination of an unlimited number of elements