Question

$100mL$ of $1.5\%(w/v)$ solution of urea is found to have an osmotic pressure of $6.0$ atm and $100mL$ 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 $\pi =CRT$ and then, calculate the total osmotic pressure.

Given, $w_1($ urea $)=1.5g$

$V_1($ urea $)=100mL$

$M_1=60g/mol$

$w_2($ cane sugar$)=3.42g$

$V_2($ cane sugar $)=100mL$

$M_2=342g/mol$

${\pi }_1($ urea $)=\frac{w_1\times 100}{M_1\times V_1}\times R\times T$

${\pi }_1($ urea $)=\frac{1.5\times 1000}{60\times 200}\times 0.082l\times 298$

${\pi }_1($ 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).