Question

Suppose you dissolve 52.2 g of Na2CO3·xH2O in enough water to make 5.00 L of solution. The final concentration of the solution was found to be 0.0366 M. Determine the integer x in the hydrate: Na2CO3·xH2O. Round your answer to the nearest integer.

Answer 1

x = 10.

Step-by-step explanation:

• Molarity is the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of Na₂CO₃·xH₂O) / (Volume of the solution (L)).

∴ no. of moles of Na₂CO₃·xH₂O = (M of Na₂CO₃·xH₂O)*(Volume of the solution (L)) = (0.0366 M)*(5.0 L) = 0.183 mol.

∵ no. of moles = mass/molar mass

∴ Molar mass of Na₂CO₃·xH₂O = (mass of Na₂CO₃·xH₂O)/(no. of moles of Na₂CO₃·xH₂O) = (52.2 g)/(0.183 mol) = 285.245 g/mol.

∵ The molar mass of Na₂CO₃·xH₂O = Molar mass of Na₂CO₃ + x (molar mass of water).

The molar mass of Na₂CO₃·xH₂O = 285.245 g/mol.

Molar mass of Na₂CO₃ = 105.9888 g/mol.

molar mass of water = 18.0 g/mol.

∴ 285.245 g/mol = 105.9888 g/mol + x (18.0 g/mol).

x (18.0 g/mol) = 285.245 g/mol - 105.9888 g/mol = 179.257 g/mol.

∴ x = (179.257 g/mol)/(18.0 g/mol) = 9.958 ≅ 10.