Question

A 999mL \rm NaCl solution is diluted to a volume of 1.29L and a concentration of 2.00\it M . What was the initial concentration?

Answer 1

Task:

A 999mL \rm NaCl solution is diluted to a volume of 1.29L and a concentration of 2.00\it M. What was the initial concentration?

Solution:

The formula for calculation of concentration is

$$C(M) = \frac{n(mol)}{V(L)}$$

C – molarity (M)

n – number of moles of substance

V – volume of solution (L)

When the solution is diluted the amount of NaCl (number of moles) remains constant. Only volume of solution changes

$$n_1 = C_1 \cdot V_1$$ $$n_2 = C_2 \cdot V_2$$

C₁ – molarity (M) of initial solution

n₁ – number of moles in initial solution

V₁ – volume of initial solution (L)

C₂ – molarity (M) of new solution

n₂ – number of moles in new solution

V₂ – volume of new solution (L)

$$n_1 = n_2$$ $$C_1 \cdot V_1 = C_2 \cdot V_2$$

The initial concentration is

$$C_1 = C_2 \cdot V_2 / V_1$$ $$C_1 = 2.00 \cdot 1.29 / 0.999 = 2.58 \, \text{M}$$

Answer: $C_1 = 2.58 \, \text{M}$