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Question: The volume of 2.8 g of carbon monoxide at \(27^{o}C\) and 0.821 atm pressure is \((R = 0.0821litatmK...

The volume of 2.8 g of carbon monoxide at 27oC27^{o}C and 0.821 atm pressure is (R=0.0821litatmK−1mol−1)(R = 0.0821litatmK^{- 1}mol^{- 1})

A

0.3 litre

B

1.5 litre

C

3 litre

D

30 litre

Answer

3 litre

Explanation

Solution

2.8 g CO =2.828=mol=0.1mol= \frac{2.8}{28} = mol = 0.1mol

PV=nRTPV = nRT

or V=nRTP=0.1×0.0821×3000.821=3litreV = \frac{nRT}{P} = \frac{0.1 \times 0.0821 \times 300}{0.821} = 3litre