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`

Step-by-step explanation:

Formula used for Elevation in boiling point :

`\Delta T_b=k_b* m`

or,

`T_b-T^o_b=i* k_b* (w_2* 1000)/(M_2* 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* (55.4g* 1000)/(60* 500g)`

`k_b=2.4^0C/m`

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