Question

A solution was prepared by dissolving acetamide (non-electrolyte), CH2CONH2 (MW =59), in 50.0g of pure water. The boiling point of the solution at 1 atm is 100.208 C. Kb of water is 0.52C kg/molHow many grams of acetamide was dissolved to yield the solution?

Answer 1

1.18 grams of acetamide were dissolved in water to produce a solution with a boiling point elevation of 0.208°C, based on the given Kb value for water.

Step-by-step explanation:

The student asked how many grams of acetamide were dissolved in water to yield a solution with a boiling point of 100.208°C, given that the Kb of water is 0.52°C kg/mol.

To find this, we will use the boiling point elevation formula, ΔTb = i * Kb * m, where ΔTb is the boiling point elevation, i is the van 't Hoff factor (which is 1 for non-electrolytes), Kb is the molal boiling point elevation constant, and m is the molality of the solution.

First, we calculate the boiling point elevation:

ΔTb = Tb(solution) - Tb(pure solvent) = 100.208°C - 100°C = 0.208°C.

Then, we solve for molality:

m = ΔTb / (Kb * i) = 0.208°C / (0.52°C kg/mol * 1) = 0.4 mol/kg.

Now, we calculate the moles of acetamide:

moles = molality * mass of solvent (kg) = 0.4 mol/kg * 0.05 kg = 0.02 mol.

Finally, we convert moles to grams:

grams = moles * molar mass = 0.02 mol * 59 g/mol = 1.18 g.

Therefore, 1.18 grams of acetamide was dissolved in the water to produce the boiling point elevation observed.