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

`0.769\ \text{M}`

Step-by-step 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=\frac{\text{Mass}}{\text{Molar mass}* \text{Volume}}\\\Rightarrow M_1=(20)/(40* 0.15)\\\Rightarrow M_1=(10)/(3)`

We have the relation

`M_1V_1=M_2V_2\\\Rightarrow M_2=(M_1V_1)/(V_2)\\\Rightarrow M_2=((10)/(3)* 15)/(65)\\\Rightarrow M_2=0.769\ \text{M}`

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