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0.45gm of a gas 1 of molecular weight 60...

0.45gm of a gas 1 of molecular weight 60 and 0.22 gm of a gas 2 of molecular weight 44 exert a total pressure of 75 cm of mercury . Calculate the partial pressure of the gas 2-

A

30 cm of Hg

B

20 cm of Hg

C

10 cm of Hg

D

40 cm of Hg

Text Solution

Verified by Experts

The correct Answer is:
A
No. of moles of gas 1 `= n_(1) = ( w_(1))/( m_(1)) =(0.45)/(60) = 0.0075`
No. of moles of gas 2 `=n_(2) = ( w_(2))/( m_(2)) =( 0.22)/( 44) = 0.0050`
Total no. of moles`= n_(1) +n_(2)`
`= 0.0075 + 0.0050 = 0.0125`
`P_(2)` partial pressure of gas 2 `= ( 00050)/( 0.0125) xx 75 = 30 cm ` of Hg
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