A certain substance X has a normal boiling point of 101.5°C and a molar boiling
ID: 1064584 • Letter: A
Question
A certain substance X has a normal boiling point of 101.5°C and a molar boiling point elevation constant Kb = 1.49 °C/m. A solution is prepared by dissolving some urea (NH2)2CO in 200 g of X. This solution boils at 104.4°C. What is the mass of urea that was dissolved in the solution?
Question 10 of 10
The normal freezing point of a certain liquid Y is -8.70°C, but when 24.17g of urea (NH2)2CO are dissolved in 550 g of Y, it is found that the solution freezes at -13.7°C instead. Use this information to calculate the molal freezing point depression constant Kf of substance Y.
A certain substance X has a normal boiling point of 101.5°C and a molar boiling point elevation constant Kb = 1.49 °C/m. A solution is prepared by dissolving some urea (NH2)2CO in 200 g of X. This solution boils at 104.4°C. What is the mass of urea that was dissolved in the solution?
A. 23.4 g B. 6.48 g C. 584 g D. 9.73 g E. 1.95 gQuestion 10 of 10
1.0 PointsThe normal freezing point of a certain liquid Y is -8.70°C, but when 24.17g of urea (NH2)2CO are dissolved in 550 g of Y, it is found that the solution freezes at -13.7°C instead. Use this information to calculate the molal freezing point depression constant Kf of substance Y.
A. 0.15°C/m B. 3.7°C/m C. 31°C/m D. 6.8°C/m E. 5.0°C/mExplanation / Answer
9.
We know that T b = Kbx m
Where
T b= elevation in boiling point
= boiling point of solution - boilinging point of pure solvent
= 104.4 - 101.5 oC
= 2.9 oC
Kb = elevation in boiling point constant of water = 1.49 oC/m
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg
= (m / 60) / 0.20 kg
= m / 12
Plug the values we get
2.9 = 1.49x(m/12)
m = 23.4 g/mol
------------------------------------------------------------
(10)
We know that T f = Kf x m
Where
T f = depression in freezing point
= freezing point of pure solvent – freezing point of solution
= -8.70 - (-13.7) oC
= 5 oC
K f = depression in freezing constant = ?
m = molality of the solution
= ( mass / Molar mass ) / weight of the solvent in Kg
= ( 24.17 / 60) / 0.550
= 0.732
Plug the values we get
Kf = 5.0 / 0.732
= 6.8 oC/m
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