Question

A student prepared a stock solution by dissolving 20.0 g of NaOH in enough water to make 150. mL of solution. She then took 15.0 mL of the stock solution and diluted it with enough water to make 65.0 mL of a final solution. What is the concentration of NaOH for the final solution

Answer 1

Answer:

$0.769\ \text{M}$

Explanation:

Mass of stock solution = 20 g

Molar mass of NaOH = 40 g/mol

Volume of stock solution = 0.150 mL

$M_2$ = Concentration of NaOH for the final solution

$V_1$ = Amount of stock solution taken = 15 mL

$V_2$ = Total volume of solution = 65 mL

Molarity is given by

$M_1=\dfrac{\text{Mass}}{\text{Molar mass}\times \text{Volume}}\\\Rightarrow M_1=\dfrac{20}{40\times 0.15}\\\Rightarrow M_1=\dfrac{10}{3}$

We have the relation

$M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{\dfrac{10}{3}\times 15}{65}\\\Rightarrow M_2=0.769\ \text{M}$

The concentration of NaOH for the final solution is $0.769\ \text{M}$.