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A solution containing 5.0 g urea per lit...

A solution containing 5.0 g urea per litre was found to be isotonic with 0.7 percent (wt./vol.) solution of an organic, non-volatile solute. Calculate the molar mass of the organic compound (molar mass of urea=60).

Text Solution

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Solutions are isotonic, therefore, `pi_(1)= pi_(2)`
`(n_(1))/(V_(1)) ST= (n_(2))/(V_(2)) ST rArr (n_(1))/(V_(1))= (n_(2))/(V_(2))` (S and T are constants)
`((w_(1))/(m_(1) xx V_(1)))_("urea") = ((w_(2))/(m_(2) xx V_(2)))_("organic") rArr (5.0)/(60xx 1000)=(0.7)/(m_(2) xx 100)`
Molar mass of the unknown organic compound, `m_(2)= (60 xx 1000 xx 0.7)/(5.0 xx 100)=84`
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