Question

A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.

Answer 1

Answer: The molal boiling point elevation constant $k_b$ of X is $2.4^0C/m$

Explanation:

Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b-T^o_b =(125.2-120.7)^0C=4.5^0C$

$k_b$ = boiling point constant  = ?

m = molality

$w_2$ = mass of solute (urea) = 55.4 g

$w_1$ = mass of solvent  X =  500 g

$M_2$ = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

$4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}$

$k_b=2.4^0C/m$

Thus the molal boiling point elevation constant $k_b$ of X is $2.4^0C/m$