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0.092 g of a compound with the molecular...

`0.092 g` of a compound with the molecular formula `C_(3)H_(8)O_(3)` on reaction with an excess of `CH_(3)MgI` gives `67.00 mL` of methane at STP. The number of active hydrogen atoms present in a molecule of the compound is :

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0.092 g of a compound with the molecular formula C_3H_8O_3 on reaction with an excess of CH_3MgI gives 67.00 mL Of methane at STP. The number of active hydrogen atoms present in a molecule of the compound

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