1.

A 20 mL,ureasolutionof 2%(w/V) is mixed with80mL of glucosesolution of 4%(w/V)at 300 K. Calculate the osmoticpressureof the solution .

Answer»

6.02 atm
1.642 atm
4.378 atm
3.01 atm

Solution :` pi = w/(MV) RT `
` = [(w/M)_("UREA") + (w/M)_("glucose")]xx (RT)/V `
Now, ` w_(urea) "in"20 mL= (2 xx 20)/(100) = 0.4 G `
`w_(" glucose") "in" 80 mL = ( 4 xx 80)/100 = 3.2 g `
` therefore= [ (0.4)/60 + (3.2)/(180)]xx ( 0.0821 xx 300 xx 1000)/(100) `
` (thereforeV = 20 + 80 = 100 mL= 100/1000 L) `
` pi = 6.02 atm `


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