'0.004 M' solution of 'Na_2 SO_4' is isotonic with '0.01 M' solution of glucose at, same temperature. The apparent degree of dissociation of 'Na_2 SO_4' is

Options

A. 0.75
B. 0.5
C. 0.25
D. 0.85

Answer

A

Explanation

We know that `pi(Na_(2)SO_4)` =iCRT =i(0.004)RT and `pi("glucose") ` =CRT =0.010RT. As the solutions are isotonic, which means they have same osmotic pressure, we have
i(0.004)RT =0.01RT
Solving, we get i=2.5. Now, consider the reaction
`Na_(2)SO_4 Leftrightarrow` Now, consider the reaction`
`Na_(2)SO_4 Leftrightarrow 2Na^+ +SO_(4)^(2-)`
Initial moles `"1 0 0"`
Moles at equilibrium `1-alpha" "2alpha" "alpha`
Total number of moles at equilibrium `=1-alpha+2alpha+alpha rArr`
`i=1+2alpha`
Therefore, `alpha=(i-1)//2=(2.5-1)//2=0.75`, So, the degree of dissociation is 75%.