1.

The mass of 112 cm^(3) " of " CH_(4) gas at STP is

Answer»

0.08g
1.6g
0.16G
0.8g

Solution :`PV = (WRT)/(M) rArr W = (MPV)/(RT)`
`= (16g MOL^(-1) xx 1"ATM" xx 0.1120"lit")/(0.082 "atm lit" k^(-1) mol^(-1) xx 273K), W= 0.089`


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