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e-lub [12.9K]
3 years ago
8

A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution

is prepared by dissolving some ammonium sulfate (NH4)2 SO4 in 600. g of X. This solution boils at 109.7 °C. Calculate the mass of (NH4)2 SO4 that was dissolved. Be sure your answer is rounded to the correct number of significant digits.
Chemistry
1 answer:
Fynjy0 [20]3 years ago
7 0

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions×\frac{1(NH_{4})_{2}SO_{4}}{3Ions} = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄×\frac{132,14g}{1mol} = <em>34,6g of (NH₄)₂SO₄</em>

<em></em>

I hope it helps!

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