InterviewSolution
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InterviewSolution
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0.092 g of a compound with the molecular C_(3)H_(8)O_(3) on reaction with an excess CH_(3)MgI gives 67.00mL of methone at STP. The number of active hydrogen atoms present in a molecule of the compound is |
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Answer» four According to the above reaction EVERY one mole of a compound contaning one active `H` atom PER molecule releases one mole of `CH_(4)` GAS at `STP`. Thus, number of moles of `CH_(4)` gas realesed per mole of compound should be an indicator of the number of active HYDROGEN atoms per molecule of compound. `n_(CH_(4))=(Vol. of CH_(4) at STP)/(22,400 mL mol^(-1))` `=0.003 mol` `n_("compound")=("mass")/("molar mass)"=(0.092g)/(92 g mol^(-1))` `=0.001 mol` This implies that `1` mol of compound releases `3` mols of `CH_(4)` gas i.e. there are `3` active hydrogen atoms present in a molecule of the compound. |
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