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A gaseous mixture contains 56 g of N2, 4...

A gaseous mixture contains 56 g of `N_2`, 44 g of `CO_2` and 16 g of `CH_4`. The total pressure of the mixture is 720 mm of Hg. The partial pressure of `CH_4` is

A

180

B

360

C

540

D

720

Text Solution

Verified by Experts

The correct Answer is:
A
Mole of `N_2 = (56)/(28) = 2 ` , Mole of `CO_2 = (44)/(44) =1 `
Mole of `CH_4 = (16)/(16) =1` , Total =2+1+1=4
`(p_(CH_4))/(p_("total"))=("moles of "CH_4)/("total moles") = (1)/(4)`
`:. p_(CH_4) = 720 xx (1)/(4) =180` mm of Hg
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