100 ml of 3% (weight/volume) urea solution and 100 ml of 6.84% (weight/volume) of cane sugar are mixed at 20°C. The solution is heated up to 27°C after mixing 100 ml water into it. Correct statements for final solution are

Urea solution is 3% (w/v)
$$\begin{gathered} \begin{array}{l}\therefore {\mathrm{n}}_{\mathrm{UREA}}=\frac{3}{60}=\frac{1}{20}\text{ moles } \\ =0.05\text{ mole in }100\mathrm{ml} \\ \mathrm{and} {\mathrm{n}}_{{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}}=\frac{6.84}{342}=0.02\text{ mole in }100\mathrm{ml} \\ \begin{array}{l}{\mathrm{V}}_{\mathrm{T}}=300\mathrm{ml}\\ \mathrm{\pi }\left(\mathrm{Urea}\right)=\frac{(0.05\times 10)\mathrm{ST}}{3}\\ =\frac{0.05\times 10\times 0.0821\times 300}{3}=4\mathrm{atm}\end{array}\end{array} \end{gathered}$$
$\begin{array}{l}\mathrm{p}\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)=\frac{0.02\times 10\times 0.0821\times 300}{3}\\ =1.64\mathrm{atm}\end{array}$
So, Total O.P. = 4 + 1.64 = 5.64 atm
Hence, (c) & (d) options are correct.