A decinormal solution of NaCl exerts an osmotic pressure of 4.6 atm. at 300 K. Calculate its degree of dissociation. `(R=0.082Latm.K^(-1)mol^(-1))`

Answer

0.87

Explanation

`pi = 0.1 xx 0.082 xx 300 = 2.46`
` i= (4.6)/(2.46) = 1.87`
m ` 1.87 = (1 + alpha)/(1) or alpha = 87% `