Question

$100 \,mL$ of $1.5 \%(w / v)$ solution of urea is found to have an osmotic pressure of $6.0$ atm and $100 \,mL$ of $3.42 \%(w / v)$ solution of cane sugar is found to have an osmotic pressure of $2.4$ atm. If the two solutions are mixed the osmotic pressure of the resulting solution in atm is
( Assume that there is no reaction between urea and cane sugar)

• A. 8.4
• B. 16.8
• C. 4.2
• D. 2.1

Answer 1

Answer: (C)

Firstly, we calculate the osmotic pressure of each solution by using the formula $\pi=C R T$ and then, calculate the total osmotic pressure.

Given, $w_{1} ($ urea $)=1.5\, g $

$V_{1} ($ urea $) =100\, mL$

$M_{1} =60\,g / mol $

$w_{2}($ cane sugar$ ) =3.42 \,g $

$V_{2}($ cane sugar $ ) =100 \,mL$

$M_{2} =342\, g / mol $

$\pi_{1} ($ urea $) =\frac{w_{1} \times 100}{M_{1} \times V_{1}} \times R \times T$

$\pi_{1} ($ urea $) =\frac{1.5 \times 1000}{60 \times 200} \times 0.082 l \times 298 $

$\pi_{1} ($ urea$ ) =3.05$ atm

Now,

$\pi_{2}($ cane sugar $)=\frac{w_{2}}{M_{2} \times V_{2}} \times 1000 \times R \times T $

$= \frac{3.42 \times 1000}{342 \times 200} \times 1000 \times 0.0821 \times 298$

Thus, the osmotic pressure of the resulting solution is

$\pi=\pi_{1}+\pi_{2}=3.05+1.22=4.27 $ atm

Thus, the nearest value is in accordance with option (c).