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balu736 [363]
2 years ago
5

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

e urea () in of . This solution boils at . Calculate the mass of that was dissolved. Round your answer to significant digits.
Chemistry
1 answer:
PSYCHO15rus [73]2 years ago
8 0

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

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