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1.12 ml of a gas is produced at STP by the action of 4.12 mg of alcohole, with methyl magnesium iodide. The molecular mass of alcohol isA. `16.0`B. `41.2`C. `82.4`D. `156.0` |
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Answer» Correct Answer - C `underset(1mol)(ROH)+CH_(3)MgBrtounderset("at S.P.T.")underset(22400mL)(CH_(4)) uarr+BrMgOR` 22400 mL of `CH_(4)` at S.T.P. = 1 mol of ROH 1.12 mL of `CH_(4)` at S.T.P. `=(1molxx1.12mL)/(22400mL)` `=5xx10^(-5)`mol `5xx10^(-6)`mol of ROH `=4.12 mg` `=4.12xx10^(-3)g` 1 mol of ROH `=(4.12xx10^(-3))/(5xx10^(-5))` `=0.824xx10^(2)=82.4g` `:.` Mol. mass of ROH = 82.4 |
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