Stock Solution of NaOH

Question

A chemist makes a stock solution of NaOH by dissolving 950.0 grams in enough water to make 150.0 liters of solution. If, later on, the chemist wants to use this stock solution to make 200.0 ml of 10.0 M NaOH, what would the chemist need to do?

Answer 1

molar mass NaOH = 39.997 g/mole

950.0 g x 1 mole/39.997 g = 23.75 moles

23.75 moles/150.0 L = 0.1583 moles/L = 0.1583 M

Since this original stock solution is already more dilute than the desired 10.0 M, one cannot readily prepare a 10.0 M solution from this original stock. If you actually wanted to do so (which no knowledgeable chemist would ever do), you would want to take 2 moles of NaOH from the original stock (this is figured from 0.2 L x 10 mol/L = 2 mol) and dissolve it in 200 ml. To do this you would need 0.1583 mol/L (x L) = 2 mol and x = 12.63 liters of the stock, evaporate it to dryness, and then redissolve the residue in 200.0 ml. Crazy!!