Question

Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO3 and water. The volume of the solution is 1.421 liters. What is the mass of solute in the solution?

Answer 1

The molarity of a solution is the number of moles of a substance divided by the volume in liters prepared.

$molarity=\frac{n}{V}$, where n is number of moles and V is the volume in liters.

In order to calculate the mass of solute we need to convert the volume and molarity to moles

$1.421 L solution \times\frac{0.0730 moles}{1 Lsolution}= 1.037 mol NaCO_3$

Now that we have moles we use the relative formula mass of NaCO₃, We have 1 Na atom, 1 C atom and 3 O atoms, thus

$M_r= (1\times 22.99) + (1\times 12.00) + (3\times 16.00)= 82.99g/mol$

$1.037 \times\frac{82.99g}{mol} = 86.1g$

Answer 2

Answer is: the mass of solute in the solution is 110 grams.

c(Na₂CO₃) = 0.730 M; molarity of sodium carbonate solution.

V(Na₂CO₃) = 1.421 L; volume of solution.

n(Na₂CO₃) = c(Na₂CO₃) · V(Na₂CO₃).

n(Na₂CO₃) = 0.73 mol/L · 1.421 L.

n(Na₂CO₃) = 1.037 mol; amount of solute.

M(Na₂CO₃) = 106 g/mol; molar mass of sodium carbonate.

m(Na₂CO₃) = n(Na₂CO₃) · M(Na₂CO₃).

m(Na₂CO₃) = 1.037 mol · 106 g/mol.

m(Na₂CO₃) = 110 g; mass of sodium carbonate.