Question

calculate the boiling point of Urea solution when 6 gram of Urea is dissolved in 200 grams of water KB for water is given 0.52 k kg per mole​

Answer 1

Answer: The boiling point of urea is 373.25 K.

Explanation:

Given that,

Weight of urea = 6 gram

Weight of water = 200 gram

$K_{b}m$of water = 0.52 k kg/mole

We know that,

$\Dela T_{b}= K_{b}m$

$T_{b}-T_{b}^{0}=K_{b}m$

$T_{b}-373 K= 0.52\ K\ kg/m\times \dfrac{6\ g\times1000}{60.06 kg/m\times 200\ g}$

$T_{b}-373 K= 2.597\ K$

$T_{b}= 2.597K+373K$

$T_{b}= 373.25 K$

$T_{b}= 373.25 K$

Hence, The boiling point of urea is 373.25 K.

Answer 2

Answer:

The BP of urea is 373.25