`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

4.67 atm

Explanation

Solute-B, Water-A
`0.85%("W/A")implies"W"_("B")=0.85"gm, V"_("sol")=100"mL"=(1)/(10)"litre" `
For `"NaNO"_(3),"n=2"`
`"I"=1+("n"-1)alpha`
`"i=1"+alpha`
`"i=1.9"`
`pi="i"xx("w"_("B")"RT")/("m"_("B")xx"V"_("sol")"(L)")`
`=1.9xx(0.85xx0.082xx300)/(85xx0.1)=4.67"atm"`