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Correct order of osmotic pressure of the following solution is
$A) 34.2 {gML}^{- 1} of glucose$
$B) 60 gm L^{- 1} of urea$
$C) 90 {gmL}^{- 1} of glucose$
$d) 58.5 {gmL}^{- 1} of NaCl$

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A<B<C<D

A<C<B<D

A<D<B<C

A<C<D<B

answer is B.

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Detailed Solution

$n_{A} = \frac{34.2}{180} = 0.2, n_{B} = \frac{60}{60} = 1, n_{C} = \frac{90}{180} = 0.5 n_{D} = \frac{58.5}{58.5} = 1$

$n_{D} = no of moles of NaCl = 1$

$NaCl \to {Na}^{+} + {Cl}^{-}$

1 mole of NaCl $\to 2 moles of ions$

$π$ bonds on no of particles.

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