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The equilibrium constant for the reaction `H_(3)BO_(3) +` glycerin `hArr (H_(3)BO_(3)` glycerine ) is 0.90. Glycerine present per litre of 0.1 M `H_(3)BO_(3)` to convert `60%` of `H_(3)BO_(3)` into `(H_(3)BO_(3)` glycerine ) is `:`A. 0.167 MB. `1.73 M`C. 0.0167 MD. 10.67 M |
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Answer» Correct Answer - B `{:(,H_(3)BO_(3),+"glycerine",hArr,H_(3)BO_(3)."glycerine"),("Initial",0.1M,,," xM"),("At eqm.",0.1(1-0.6)M,,,(x-0.06)M " "0.06M):}` `K=([H_(3)BO_(3)."glycerine"])/([H_(3)BO_(3)]["glycerine"])` `=0.9=(0.06)/((0.04)(x-0.06))` `(9)/(10)xx(4)/(6)=(1)/(x-0.06)` `x-0.06=1.67` `x=1.67+0.06` `x=1.73M` |
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