1.

The number of moles of hydroxide (HO^(-))ion in 0.3 litre of 0.005 M solution of Ba(OH)_(2) is

Answer»

0.0075
0.0015
0.003
0.005

Solution :0.3L of 0.005 M `Ba(OH)_(2)` SOL as`M=n/V(L)`
`:. N = 0.005 M Ba(OH)_(2)` sol as `M=n/(V(L))`
`:.n = 0.005 XX 0.3` mol = 0.0015 mol `Ba(OH)_(2)` `(Ba^(2+)2OH^(-))`
mol of `OH^(-)` = `2 xx 0.0015 = 0.0030` moles


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