Question

a saturated solution of nacl (dissolved in water) has a concentration of 5.79 m at a given temperature. the solution has a volume of 3 l. if the water were to evaporate until the volume of the solution is 453 ml (and then stirred), what would the concentration of the new solution be?

Answer 1

The concentration of the new solution would be 38.34 M.

Step-by-step explanation:

Given that the saturated solution of NaCl has a concentration of 5.79 M and a volume of 3 L, we can find the amount of NaCl present in the solution by multiplying the concentration and volume:

Amount of NaCl = Concentration x Volume

Amount of NaCl = 5.79 M x 3 L = 17.37 moles

When the solution is evaporated to a volume of 453 mL (0.453 L), the amount of NaCl remains the same:

Amount of NaCl = 17.37 moles

The new concentration can be found by dividing the amount of NaCl by the new volume:

New Concentration = Amount of NaCl / New Volume

New Concentration = 17.37 moles / 0.453 L = 38.34 M

Therefore, the concentration of the new solution would be 38.34 M.