Question

14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and distilled water, when mixed, will result in a solution that is approximately 1 M NaOH?

Answer 1

The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

$\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M$

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock solution.

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution.

From data (A) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 20mL=1M\times V_2\\\\V_2=100mL$

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 20mL=1M\times V_2\\\\V_2=100mL$

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

$M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 60mL=1M\times V_2\\\\V_2=300mL$

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 60mL=1M\times V_2\\\\V_2=300mL$

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.