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`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)`) |
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Answer» Correct Answer - 4.67 atm 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"` |
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