[audienne]

 A Solution of potassium hydroxide has a concentration of 5.0Gg/L. 150 ml of this solution is diluted with water to 3.0L. What is the concentration of this new solution


Molar mass of KOH solute = 56g ( K 39g + O 16g +H 1g)

5.0g of KOH = 5g/56gx 1 mol = 0.089 mole
                         

Molarity of original solution = 0.089 mol/L

No of mole present in 150ml of KOH solution = 0.15L x 0.089mol/L = 0.01335 mole

By definition, number of moles present in the more dilute solution equals the number of moles of solute that must be present in the more concentrated stock

Molarity  = Number of mole
                     Volume

Molarity of 3.0L = 0.01335 mole /3L = 0.00445 M/L
                                           

Concentration ( w/v) = 0.00445x56g/L = 0.2492g/L

[Borek]

Your final result is (almost) OK - you just should remember to not use too much digits and write it as 0.25 g/L. But...

There is no need to use moles concept to solve this question. At all.

You start with the 5.0g/L solution.

You take 150mL, or 0.15L of that - so your solution contains 0.15*5.0 = 0.75g of NaOH.

You dilute it in 3.0 liters, so your final concentration is 0.75g/3.L = 0.25 g/L.

Much simpler and much faster

[audienne]

Thanks heaps!