Question

A certain substance X condenses at a temperature of 123.3 degree C . But if a 650. g sample of X is prepared with 24.6 g of urea ((NH2)2 CO) dissolved in it, the sample is found to have a condensation point of 124.3 degree C instead. Calculate the molal boiling point elevation constant Kb of X.Round your answer to 2 significant digits.

Answer 1

Answer:

The molal boiling point elevation constant is 1.59 ≈  1.6 $Kkgmol^{-1}$

Explanation:

To solve this question , we will make use of the equation ,

Δ$T_{b} = i*K_{b} *m$

where ,

• Δ$T_{b}$ is the change in boiling point of the substance $X$ ( °$C$ or $K$)

• $i$ is the Vant Hoff Factor which = 1 in this case ( no unit )

• $K_{b}$ is the mola boiling point elevation constant of X ( $Kkgmol^{-1}$)

• $m$ is the molality of the solution which has $(NH_{2})_{2} CO$ as the solute and  $X$ as the solution ($molkg^{-1}$)

1. Δ$T_{b}$ = $124.3 -123.3 = 1$ °$C$ or $K$;
2. $i$=1;
3. $m$= $\frac{moles of solute}{weight of solvent(kg)}$$molkg^{-1}$

           ∴ $m = \frac{\frac{24.6}{60} }{\frac{650}{1000} }$

• as the weight of $(NH_{2})_{2} CO$ is $60g$ and thus number of moles = $\frac{24.6}{60}$

• and the weight of solvent in $kg$ is $\frac{650}{1000}$

    4. $K_{b}$ ⇒ ?

∴

$1=1*K_{b} *\frac{\frac{24.6}{60} }{\frac{650}{1000} }$

⇒ $K_{b}$ = $1.59$ ≈ 1.6 $Kkgmol^{-1}$