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The molecular weight of a gas is 44 amu....

The molecular weight of a gas is 44 amu. The volume occupied by 2.2 g of this gas at `0^(@) C ` and 2 atm pressure will be

A

0.56 L

B

1.2L

C

2.4L

D

5.6 L

Text Solution

Verified by Experts

The correct Answer is:
A
Number of moles of gas `= ( 22)/( 44) = (1)/(20) ` mole
One mole of a gas occupies 22.4 L at NTP
`rho V = n RT `
`rArr ( rho_(1) V_(1))/( rho_(2) V_(2)) = (n_(1))/( n_(2)) rArr V_(2) = ( rho_(1))/( rho_(2)) xx ( n_(1))/( n_(2)) V_(1) = ( 22 . 4)/( 20 xx 2)`
`rArr V_(2) = 0 . 56 L`
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