Question

Suppose that 0.100 L of a 2.5 M NaCl stock solution is obtained. This volume is then diluted to a final volume of 0.500 L by adding 400 mL of water a) Determine the numbber of moles of NaCl in the 0.100 L ____ mol b ) Determine the number of moles of NaCl present once the solution was diluted to 0.500 L _____mol c) Determine the final concentration of the diluted solution _______M

Answer 1

(a) Number of moles , n = Molarity x volume in L

= 2.5M x 0.100 L

= 0.25 mol

(b) According to law of dilution   MV = M'V'

Where M = Molarity of stock = 2.5 M

V = Volume of the stock = 0.100L

M' = Molarity of dilute solution = ?

V' = Volume of the dilute solution = 0.500L

Plug the values we get   , M' = MV /V' = 0.5 M

So number of moles of NaCl , n' = Molarity x volueme in L

= 0.5M x 0.500L

= 0.5 mol

(c) The final concentration of the diluted solution is 0.5 M