A 914 mL NaCl solution is diluted to a volume of 122 L and a

A 914 mL NaCl solution is diluted to a volume of 1.22 L and a concentration of 4.00 M . What was the initial concentration?

Solution

Let the initial concentration of NaCl solution is xM.

M is molarity of solution. Molarity is no of moles present in one litre of solution mixture.

914 ml = 0.914 L of initial NaCl solution= V₁

Moles of NaCl present intially = x * V₁

= 0.914 * x

Additional water added = 1.22- V₁

= 1.22- 0.914 = 0.306 L

Initial moles of NaCl will be same as final mole of NaCl because only water is added.

so 0.914 * x = 1.22 * 4

it gives x= 5.34 M