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The osmotic pressure of a solution containing `40 g` of solute `("molecular mass 246")` per litre at `27^(@)C` is `(R=0.0822 atm L mol^(-1))`A. `3.0 atm`B. `4.0 atm`C. `2.0 atm`D. `1.0 atm` |
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Answer» Correct Answer - B `pi=CRT=((W_(2)/(Mw_(2)))RT)/V` Given`W_(2)=40 g` `Mw_(2) = 246` `T=27^(@)C = 300 K` `V= 1 L` Substituting all the values , we get `pi=(40/246 )xx 0.082 xx 300 =4 atm` |
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