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.

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×[tex]\frac{1(NH_{4})_{2}SO_{4}}{3Ions}[/tex] = 0,262 moles of (NH₄)₂SO₄

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

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

I hope it helps!