1.

A 0.004 M Solution of Na_(2)SO_(4) is isotonic with a 0.010 M solution of glucose at the temperature. The apparent degree of dissociation of Na_(2)SO_(4) is

Answer»

`25%`
`50%`
`75%`
`85%`

Solution :`pi(Na_(2)SO_(4))=iCRT=i(0.004)RT`
`pi"(Glucose) = CRT = 0.010 RT"`
As solution are isotonic, I (0.004) RT = 0.01 RT.
This givesI = 2.5
`{:("Now",Na_(2)SO_(4),hArr,2Na^(+),+,SO_(4)^(2-)),(,"1 MOLE",,0,,0),(,1-alpha,,2ALPHA,,alpha","):}`
`"Total "=1+2alpha`
`therefore""i=1+2alpha`
`"or"alpha=(i-1)/(2)=(2.5-1)/(2)=0.75=75%`.


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