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Calculate the osmotic pressure of 0.2 M glucose solution at 300 K. `(R=8.314 J mol^(-1) K^(-1))` |
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Answer» Correct Answer - Osmotic pressure `=4.988 xx 10^(5) Nm^(-2)` Given : Concentration of the solution `=c= 0.2 M` `=0.2 "mol" dm^(-2)` `=0.2 xx 10^(3) "mol" m^(-3)` Temperature `= T= 300 K` `R= 8.314 j mol^(-1) K^(-1)` The osmotic pressure `pi` is given by `pi` = CRT `=0.2 xx 10^(3) xx 8.314 xx 300` `=4.988 xx 10^(5) Nm^(-2)` (or pa)` |
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