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0.092 g of a compound with the molecular...

0.092 g of a compound with the molecular formula `C_3H_8O_3` on reaction with an excess of `CH_3`Mgl gives 67.00 mL of methane at STP. The number of active hydrogen atoms presentin a molecule of the compound is :

A

one

B

two

C

three

D

four

Text Solution

Verified by Experts

92g ofa compound react with excess of `CH_3 MgI` to give `x xx 22400` mL of `CH_4` at STP.
1 g of a compound react with excess of `CH_3` MgI to give `(x xx 22400)/(92)`
0.092 of a compound react with excess of `CH_3` MgI to give `(x xx 22400)/(92) xx 0.092`
`(x xx 22400 xx 0.092)/(92) = (67 xx 92 xx 1000)/(22400 xx 92) = (670)/(224) = 2.99 = 3`
`x rArr` number ofactive hydrogen in compound.
active `H rArr 3`
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