When `0.004 M Na_(2)SO_(4)` is an isotonic acid with `0.01 M `glucose, the degree of dissociation of `Na_(2)SO_(4)` is

Options

A. `75%`
B. `85%`
C. `50%`
D. `25%`

Answer

1

Explanation

Isotonic means their effective molarition are equal:
`(i C)_(Na_(2)SO_(4))=(I C)_("glucose")`
`i_(Na_(2)SO_(4))=((I C)_("glucose"))/C_(Na_(2)SO_(4)`
`=((1)(0.01 M))/((0.004 M))`
`=2.5`
`{:(,Na_(2)SO_(4)(aq)hArr,2Na^(+)(aq),+,SO_(4)^(2-)(aq)),("Moles before dissociation",1mol,0mol,,0mol),("Moles before dissociation",(1-alpha)mol,2alphamol,,alphamol):}`
`i=("Total moles of particles after dissociation")/("Total moles of particles before dissociation")`
`i=((1-alpha)+(2alpha)+(alpha))/1`
`i=1+2alpha`
`alpha=(i-1)/2=(2.5-1)/2`
`=0.75`
Thus, percent dissociation`=0.75xx1.007`.