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4.4 g of CO(2) and 2.24 "litre" of H(2) ...

`4.4 g of CO_(2)` and `2.24 "litre" of H_(2)` at `STP` are mixed in a container. The total number of molecules present in the container will be:

A

`9.9`

B

`0.099`

C

`0.001`

D

`1.00`

Text Solution

Verified by Experts

The correct Answer is:
B
`PV=nRT, P=1 atm, T=273 K, V = 2.24 L, R=0.0821 L atm K^(-1)mol^(-1)`
`n=(1xx2.24)/(0.0821xx273)=0.099`
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