This question is incomplete, the complete question is;
A certain liquid X has a normal boiling point of 150.4 °C and a molar boiling point elevation constant kb is 0.60 °Ckgmol⁻¹.
A solution is prepared by dissolving some urea (NH22CO) in 750 g of X. This solution boils at 150.9 °C . Calculate the mass of urea that was dissolved. Round your answer to 3 significant digits.
Answer:
the mass of urea that was dissolved is 37.5 g
Explanation:
Given the data in the question;
normal boiling point of X; Tb⁰ = 150.4 °C
boiling point of solution Tb = 150.9 °C
Change in boiling point Δt = Tb - Tb⁰ = 150.9 °C - 150.4 °C = 0.5 °C
Kb = 0.6 °C.kg.mol⁻¹
V = 750 g
Now, we know that
Δt = Kb × molality
so
0.5 = 0.6 × molality
molality = 0.5 / 0.6
molality = 0.833
we know that molar mass of urea is 60 g/mol
so
molality = mass × 1000 / molar mass × volume( g )
we substitute
0.833 = ( mass × 1000 ) / ( 60 × 750 )
0.833 = ( mass × 1000 ) / 45000
0.833 × 45000 = mass × 1000
mass = ( 0.833 × 45000 ) / 1000
mass = 37485 / 1000
mass = 37.485 ≈ 37.5 g { 3 significance figure }
Therefore, the mass of urea that was dissolved is 37.5 g
