1.

The volume occupied by 0.32 g of a gas at S.T.R is 224 mL. Calculate the molecular mass of the gas.

Answer»

SOLUTION :`therefore` 224 mL of the gas at S.T.P WEIGH = 0.32 G
`therefore 22400 mL` of the gas at S.T.P. will weigh
`=0.32/224 xx 22400 = 32 g`
HENCE, the molecular mass of the gas = 32 amu


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