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

The molal boiling point elevation constant (Kb) of substance X is 4.1.

Step-by-step explanation:

The molal boiling point elevation constant (Kb) can be calculated using the formula: ΔT = Kb × m

Where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.

In this case, the change in boiling point (ΔT) is 1 degree C (124.3 - 123.3), the molality (m) can be calculated by dividing the molal mass of urea by the mass of the solvent water, which gives a value of 0.0246 kg urea / 0.100 kg water = 0.246 mol/kg, and the formula becomes: 1 = Kb × 0.246

Now, rearrange the equation to solve for Kb: Kb = 1 / 0.246 = 4.07

Rounding to 2 significant digits, the molal boiling point elevation constant Kb of substance X is 4.1.

Answer 2

The molal boiling point elevation constant is 1.59 ≈ 1.6
`Kkgmol^(-1)`

Step-by-step 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`=
   `(moles of solute)/(weight of solvent(kg))`
   `molkg^(-1)`

∴
`m = ((24.6)/(60) )/((650)/(1000) )`

• as the weight of
  `(NH_(2))_(2) CO` is
  `60g` and thus number of moles =
  `(24.6)/(60)`

• and the weight of solvent in
  `kg` is
  `(650)/(1000)`

4.
`K_(b)` ⇒ ?

∴

`1=1*K_(b) *((24.6)/(60) )/((650)/(1000) )`

⇒
`K_(b)` =
`1.59` ≈ 1.6
`Kkgmol^(-1)`