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 `:`

Options

A. 0.167 M
B. `1.73 M`
C. 0.0167 M
D. 10.67 M

Answer

B

Explanation

`{:(,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`