$22.4$ litres of a gas weighs $70$ g at S.T.P. Calculate the weight (gm) of the gas if it occupies a volume of $20$ litres at $27^{o}$ C and $700$ mm Hg of pressure.

Open in App

Solution

Let us consider the problem:
$P V = n R T - - - - - 1$
Given that:
$R = 0.082$
$P = \frac{700}{760} = 0.921 a t m$
$T = 273 + 27 = 300$
Put the all value in equation 1
$0.921 \times 20 = n \times 0.082 \times 300$
$n = 0.748 m o l e s$
$1$ mole contains $70 g$
$0.748 c o n t a i n s - - - - - ?$
$0.748 \times \frac{70}{1} = 52.414$

Suggest Corrections

Similar questions

22.4 litres of a gas weighs 70 g at S.T.P. Calculate the weight of the gas if it occupies a volume of 20 litres at $27^{\circ} C$ and 700 mm Hg of pressure.

Q. Calculate the volume occupied by

2 g

of hydrogen at

27^{\circ} C

and

4

atmosphere pressure, if at S.T.P. it occupies

22.4

litres.

Q. 22.4 litres of a gas at S.T.P. weigh 16 g. Identify the gas.

Q. The molecular weight of 2.6 gm of a gas which occupies 2.24 litres at

0^{o} C

& 760 mm pressure is :

Join BYJU'S Learning Program

22.4 litres of a gas weighs 70g at S.T.P. Calculate the weight (gm) of the gas if it occupies a volume of 20 litres at 27oC and 700mm Hg of pressure.

Let us consider the problem:PV=nRT−−−−−1Given that:R=0.082P=700760=0.921 atmT=273+27=300 Put the all value in equation 10.921×20=n×0.082×300n=0.748 moles1 mole contains 70g0.748 contains −−−−−?0.748×701=52.414

$22.4$ litres of a gas weighs $70$ g at S.T.P. Calculate the weight (gm) of the gas if it occupies a volume of $20$ litres at $27^{o}$ C and $700$ mm Hg of pressure.