Question

Osmotic pressure of solution containing 0.6 g urea
and 3.42 g sugar in 100 ml at 27°C
(1) 492 atm
(2) 4.92 atm
(3) 49.2 atm
(4) 28.1 atm​

Answer 1

Answer:

(2) 4.92 atm is the right answer.

Explanation:

Refer attachment for explanation!!!

Answer 2

Given,

mass of urea($NH_2CONH_2$) $= 0.6 g$

gram molecular weight of urea = $60 g/mole$

mass of sugar($C_{12}H_{22}O_{11}$)$= 3.42 g$

gram molecular weight of sugar $= 342 g/mol$

Now, we know the formula for osmotic pressure as

$\pi_{osmotic} = \frac{n_{urea}+n_{sugar}}{v}*R*T \\ \\\pi = \frac{\frac{0.6}{60}+\frac{3.42}{342} }{\frac{100}{1000} } *0.082*300$

Here,  $n =$ no. of moles of the substances is the solution.

          $R = 0.082 \frac{atm.lit}{mol}$

          $v = \frac{100}{1000} lit$

          $T = 300 K$

Therefore,

$\pi = \frac{0.02}{0.1}*0.082*300\\ \\ \pi = 4.92 atm$

∴The osmotic pressure of the solution is $\pi = 4.92 atm$

correct option is $(2). 4.92 atm$

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