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0.037g of an alcohol, R-OH was added to `CH_(3)MgBr` and the gas evolved measured 11.2 mL at STP. The Molecular mass of R-OH will be .A. 47B. 79C. 74D. 77 |
Answer» Correct Answer - C [Hint] `{:(R-OH,+,CH_(3)MgBr,rarr,CH_(4),+,Mg(Br)OR),(1 mol,,,,1 mol,,),(,,,,or,,),(,,,,22400 "mL at STP",,):}` ltBRgt 11.2 Ml `ch_(4)` at STP is formed by 0.037 g ROH `:. ` 22400 mL `CH_(4)` at STP will be formed by `(0.037)/(11.2)xx22400`g ROH =74 g ROH `:. `Molar mass of R-OH `=74` |
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