If you add 90 mL of 0.15M stock solution, to 10 mL of water, what is the new concentration?

0,15 moles of NaOHin1000ml

x moles of NaOHin100ml

x = 0,015 moles of NaOH

final volume = 150ml

0,015 moles of NaOH---in150ml

x moles of NaOHin1000ml

x = 0,1 moles of NaOH

 0,1mol/dm³  (molarity)

the answer is either b or c