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Calculate the osmotic pressure at 273 K of a 5% solution of urea (Mol. Mass = 60). (R = 0.0821 litre atm/degree/mole). |
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Answer» Solution :`5%` solution of urea means that it contains 5 g of urea per `100cm^(3)` of the solution, i.e., `w_(2)=5g, V=100 cm^(3)=(1)/(10)" litre = 0.1 litre."` `M_(2)-"60 g MOL"^(-1), R="0.0821 L atm K"^(-1)"mol"^(-1), T= 273 K` `therefore""pi=(n)/(V)RT=(w_(2))/(M_(2))xx(1)/(V)RT=(w_(2)RT)/(M_(2)V)=((5g)("0.08211 L atm K"^(-1)"mol"^(-1))(273 K))/(("60 g mol"^(-1))("0.1 L"))="18.68 atm."` |
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