$100 \mathrm{~mL}$ of $1.5 \%(\mathrm{w} / \mathrm{v})$ solution of urea is found to have an osmotic…
$100 \mathrm{~mL}$ of $1.5 \%(\mathrm{w} / \mathrm{v})$ solution of urea is found to have an osmotic pressure of $6.0 \mathrm{~atm}$ and $100 \mathrm{~mL}$ of $3.42 \%(w / v)$ solution of cane sugar is found to have an osmotic pressure of $2.4 \mathrm{~atm}$. 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)
8.4
16.8
4.2
2.1
Solution
(c) Key Idea Firstly, we calculate the osmotic pressure of each solution by using the formula $\pi=C R T$ and then, calculate the total osmotic pressure.
$
\begin{aligned}
\text { Given, } w_1(\text { urea }) & =1.5 \mathrm{~g} \\
V_1(\text { urea }) & =100 \mathrm{~mL} \\
M_1 & =60 \mathrm{~g} / \mathrm{mol} \\
w_2(\text { cane sugar }) & =3.42 \mathrm{~g} \\
V_2(\text { cane sugar }) & =100 \mathrm{~mL} \\
M_2 & =342 \mathrm{~g} / \mathrm{mol} \\
\pi_1(\text { urea }) & =\frac{w_1 \times 100}{M_1 \times V_1} \times R \times T \\
\pi_1(\text { urea }) & =\frac{1.5 \times 1000}{60 \times 200} \times 0.0821 \times 298 \\
\pi_1(\text { urea }) & =3.05 \mathrm{~atm}
\end{aligned}
$
Now,
$
\begin{aligned}
& \pi_2(\text { 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
\end{aligned}
$
Thus, the osmotic pressure of the resulting solution is
$
\pi=\pi_1+\pi_2=3.05+1.22=4.27 \mathrm{~atm}
$
Thus, the nearest value is in accordance with option (c)