5.2 g of polyhydric alcohol was treated with an excess of methyl magnesium bromide to produce 3.36 litre of CH 4 at STP. Calculate number of OH groups present in polyhydric alcohol (molar mass of alcohol =104 g mol -1 )

Question

5.2 g of polyhydric alcohol was treated with an excess of methyl magnesium bromide to produce 3.36 litre of CH 4 at STP. Calculate number of OH groups present in polyhydric alcohol (molar mass of alcohol =104 g mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Number of moles of alcohol =(5.2/104)=(1/20) Number of moles of CH 4 produced =(3.36/(22.4 lt ))=(3/20) So number of - OH groups presents =3

Q. $5.2 g$ of polyhydric alcohol was treated with an excess of methyl magnesium bromide to produce $3.36$ litre of $C H_{4}$ at STP. Calculate number of OH groups present in polyhydric alcohol (molar mass of alcohol $= 104 g m o l^{- 1}$ )

Number of moles of alcohol $= \frac{5.2}{104} = \frac{1}{20}$
Number of moles of $C H_{4}$ produced
$= \frac{3.36}{( 22.4 lt )} = \frac{3}{20}$
So number of $- O H$ groups presents $= 3$

Q. 5.2g of polyhydric alcohol was treated with an excess of methyl magnesium bromide to produce 3.36 litre of CH4​ at STP. Calculate number of OH groups present in polyhydric alcohol (molar mass of alcohol =104gmol−1 )

Number of moles of alcohol =1045.2​=201​ Number of moles of CH4​ produced =(22.4lt)3.36​=203​ So number of −OH groups presents =3

Q. $5.2 g$ of polyhydric alcohol was treated with an excess of methyl magnesium bromide to produce $3.36$ litre of $C H_{4}$ at STP. Calculate number of OH groups present in polyhydric alcohol (molar mass of alcohol $= 104 g m o l^{- 1}$ )

Number of moles of alcohol $= \frac{5.2}{104} = \frac{1}{20}$
Number of moles of $C H_{4}$ produced
$= \frac{3.36}{( 22.4 lt )} = \frac{3}{20}$
So number of $- O H$ groups presents $= 3$