Question

(a) Sample of NaOH weighing 0.38 is dissolved in water and the solution is made to 50.0 mL in a volumetric flask. What is the molarity of the resulting solution ?
(b) How many moles of NaOH are contained in 27 mL of 0.15 M NaOH solution ?

Answer 1

(a) Molarity of solution (M) `= ("No. of moles of NaOH")/("Volume of solution in litres")`
`=((0.38//40"g mol"^(-1)))/((50//100L))=(0.38xx1000)/(40xx50)("mol L"^(-1))`
`=0.19 mol L^(-1) = 0.19 M`
(b) Molarity of solution (M) `=0.15 M = 0.15 mol L^(-1)`
Volume of solution `= 27 mL = (27)/(1000)=0.027 L`
Molarity of solution `= ("No. of moles of NaOH")/("Volume of solution in Litres")`
`(0.15 "mol L"^(-1))=("No. of moles of NaOH")/("0.027 L")`
`:.` No. of moles of NaOH `=(0.15 "mol L"^(-1))xx(0.027 L)=4.05 xx 10^(-3)` mol.