The Temperature at which 28 g of N(2) wi...

The Temperature at which 28 g of `N_(2)` will occupy a volume of 10.0 L at 2.46 atm is

A

299.6 K

B

`0^(@) C`

C

`273 K`

D

`10^(@) C`

Text Solution

Verified by Experts

The correct Answer is:

A

`28 g N_(2) = 1` mole of `N_(2)`
Applying pV = nRT
`2.46 xx 10 = 1 xx 0.0821 xx T` or T = 299.6 K

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