Question

20 g of an ideal gas contains only atoms of S and O occupies 5.6 L at 1 atm and 273k. what is the molar wt. of gas? please answer step by step. thx

Answer 1

thus the ans is 80 gm

Answer 2

Answer : The molar mass of the gas is, 80.0475 g/mole

Solution : Given,

Mass of gas = 20 g

Pressure of gas = 1 atm

Volume of gas = 5.6 L

Temperature of gas = 273 K

Using ideal gas law,

$PV=nRT\\\\PV=\frac{w}{M}\times RT\\\\M=\frac{wRT}{PV}$

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

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

$M=\frac{wRT}{PV}$

$M=\frac{(20g)\times (0.0821Latm/moleK)\times (273K)}{(1atm)\times (5.6L)}=80.0475g/mole$

Therefore, the molar mass of the gas is, 80.0475 g/mole