Question

If 16 grams of CH4 and 4 grams of H2 are mixed and kept at 760 mm Hg pressure at 0 degrees Celsius then volume occupied by the mixture would be ​

Answer 1

Answer:

Explanation:

n1(ch4) =  16/16 = 1

n2(H2) = 4/2 = 2

n(total) = n1+n2 = 3

V = nRT/P

 =  3 *0.083 *273/1                .............1atm = 760mm of hg

 =  67L