1.

0.85% aqueous solution of NaNO_(3) is apparently 90% dissociated at 27^(@)C. Calculate its osmotic pressure. (R = 0.0821 atm K^(-1) mol^(-1))

Answer»

Solution :Volume of `NaNO_(3)` solution
`=((100g))/((1"G cm"^(-3)))=100cm^(3)=0.1 L`
`NaON_(3)` dissociates as :
`NaON_(3)toNa^(+)+NO_(3)^(-)`
`alpha=((i-1))/((n-1))i.e.0.9(i-1)/(2-1)`
i-1+0.9=1.9
`M_(B)=23+14+48=85" g mol"^(-1)`
`pi=W_(B)/M_(B)XX(RxxTxxi)/V`
`=((0.85))/((85" g mol"^(-1)))`
`xx((0.082" L ATM K"^(-1)"mol"^(-1))xx(300L)xx(1.0))/((0.1 L))=4.67~~5`


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