Question

How do you prepare 250.00 mL of 3.00 M HCl solution starting with a concentrated HCl solution which is 36.0% HCl by mass and has a density of 1.18 g/mL?

Answer 1

Answer: 64.1 ml of concentrated
`HCl` is taken and (250-64.4)= 185.6 ml f water is added to make 250.00 mL of 3.00 M HCl .

Explanation:

Given : 36 g of
`HCl` is dissolved in 100 g of solution.

Density of solution = 1.18 g/ml

Thus volume of solution =
`(mass)/(density)=(100)/(1.18)=84.7ml`

`Molarity=(n* 1000)/(V_s)`

where,

n= moles of solute

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}=(36g)/(36.5g/mol)=0.99moles`

`V_s` = volume of solution

`Molarity=(0.99* 1000)/(84.7)=11.7M`

According to the dilution law,

`M_1V_1=M_2V_2`

where,

`M_1` = molarity of stock
`HCl` solution = 11.7M

`V_1` = volume of stock
`HCl` solution = V ml

`M_2` = molarity of diluted
`HCl` solution = 3.00 M

`V_2` = volume of diluted
`HCl` solution = 250.0 ml

`11.7M* V=3.00* 250.0`

`V=64.1ml`

Thus 64.1 ml of stock
`HCl` is taken and (250-64.4)= 185.6 ml of water is added to make 250.00 mL of 3.00 M HCl .