Calculate the osmotic pressure of a solution obtained by mixing `100 cm^(3)` of `1.5%` solution of urea (mol. Mass=60) and `100 cm^(3)` of `3.42%` solution by cane sugar (mol. Mass = 342) at `20^(@)C`. (R=0.082 litre atm/deg/mole)

Calculate the osmotic pressure of a solution obtained by mixing `100 c

1.	Calculate the osmotic pressure of a solution obtained by mixing `100 cm^(3)` of `1.5%` solution of urea (mol. Mass=60) and `100 cm^(3)` of `3.42%` solution by cane sugar (mol. Mass = 342) at `20^(@)C`. (R=0.082 litre atm/deg/mole)
Answer» After mole total volume of the solution `=100 + 100=200 cm^(3)` Osmotic pressure due to the urea in the solution. `1.5 g` of urea which was present originally in `100 cm^(3)` is now present in `200 cm^(3)`,i.e., in the final solution. `Mw_(1)=Mw_("urea")=(W_(2) xx 1000)/(Mw_(2) xx vol. of sol.)=(1.5 xx 1000)/(60 xx 200)=0.125` `Mw_(2)=Mw_(sugar)=(3.42 xx 1000)/(342 xx 200)=0.05` `pi=(Mw_(1)+Mw_(2))RT` `=(0.125 + 0.05) xx0.82 xx 293 =4.2 atm`

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Calculate the osmotic pressure of a solution obtained by mixing `100 cm^(3)` of `1.5%` solution of urea (mol. Mass=60) and `100 cm^(3)` of `3.42%` solution by cane sugar (mol. Mass = 342) at `20^(@)C`. (R=0.082 litre atm/deg/mole)

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