Example Problem Describe the preparation of 1.0x102 mL of 6.0 M HCl from a concentrated solution that has a specific gravity of 1.18 and is 37% (w/w) HCl (36.5 g/mol)

Question

asked May 5, 2020 213k views

1 Answer

Chris Hines · Answer 1 · 2020-05-10T12:19:24+0000

Answer: 15.3 ml of water is to be added to 7.1 M solution.

Step-by-step explanation:

The relationship between specific gravity and density of a substance is given as:

$\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}$

Specific gravity of
HCl = 1.18

Density of water =
1.00g/ml

Putting values in above equation we get:

$1.18=\frac{\text{Density of solution}}{1.00g/ml}\\\\\text{Density of solution}=1.18g/ml$

Given : 37 g of HCl is present in 100 g of solution

Volume of concentrated HCl solution=
$\frac{\text{Mass of solution}}{\text{Density of solution}}=(100g)/(1.18g/ml)=84.7ml$

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock
HCl solution = ?

M_2 = molarity of diluted
HCl solution= 6.0 M

V_2 = volume of stock
HCl solution = 84.7 ml

V_2 = volume of diluted
HCl solution =
1.0* 10^2ml

M_1* 84.7=6.0* 1.0* 10^2

M_1=7.1

Therefore, the concentration of stock
HCl solution is 7.1 M

$Molarity=\frac{moles}{\text {volume in L}}$

7.1=(moles)/(0.0847L)

moles=7.1* 0.0847=0.60

Mass of
$HCl=moles* {\text{Molar mass}}=0.60mol* 36.5g/mol=21.9g$

To prepare 7.1 M solution , we need to add 21.9 g of HCl to 84.7 ml of water.

To prepare
of 6.0 M HCl from a concentrated solution , (100-84.7) = 15.3 ml of water is to be added to 7.1 M solution.