Question

If 0.2 mol urea is added into 1000g water then boiling point of the solution is (kb (water) =0.52K/molal)

Answer 1

Answer:ch2cl-ch=ch-ch2cl

Explanation

Answer 2

Answer:

The boiling point of the urea solution is equal to $373.104 \mathrm{~K} .$

Therefore, the option (1) is correct.

Explanation:

Given:

The molal elevation constant $K_{b}=0.52 \mathrm{~K} \mathrm{kgmol}^{-1}$

The number of moles of urea $=0.2 \mathrm{~mol}$

The mass of water (solvent) $1000 \mathrm{~g}=1 \mathrm{~kg}$

We know,

Molality of solution $=(\frac{no. of moles of solute}{ mass of solvent} )$

Molality $=\frac{0.2 \mathrm{~mol}}{1 \mathrm{~kg}}=0.2 \mathrm{molkg}^{-1}$

The elevation in boiling point is given by : $$\Delta T_{b}=m * K_{b} \Delta^{T_{b}}=\left(0.2 \mathrm{molkg}^{-1}\right)\left(0.52 \mathrm{Kkgmol}^{-1}\right)$

$\Delta T_{b}=0.104 K$

We know, pure water has boiling point, $T_{b}^{o}=373 K$

$T_{b}-T_{b}^{o}=0.104 K$

$T_{b}=T_{b}^{o}+0.104 K$

$\begin{aligned}&T_{b}=373 K+0.104 K \\&T_{b}=373.104 K\end{aligned}$

Therefore, boiling point of solution is $373.104 \mathrm{~K} .$