Respuesta :
Answer: 15.3 ml of water is to be added to 7.1 M solution.
Explanation:
The relationship between specific gravity and density of a substance is given as:
[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]
Specific gravity of [tex]HCl[/tex] = 1.18
Density of water = [tex]1.00g/ml[/tex]
Putting values in above equation we get:
[tex]1.18=\frac{\text{Density of solution}}{1.00g/ml}\\\\\text{Density of solution}=1.18g/ml[/tex]
Given : 37 g of HCl is present in 100 g of solution
Volume of concentrated HCl solution=[tex]\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.18g/ml}=84.7ml[/tex]
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]HCl[/tex] solution = ?
[tex]M_2[/tex] = molarity of diluted [tex]HCl[/tex] solution= 6.0 M
[tex]V_2[/tex] = volume of stock [tex]HCl[/tex] solution = 84.7 ml
[tex]V_2[/tex] = volume of diluted [tex]HCl[/tex] solution = [tex]1.0\times 10^2ml[/tex]
[tex]M_1\times 84.7=6.0\times 1.0\times 10^2[/tex]
[tex]M_1=7.1[/tex]
Therefore, the concentration of stock [tex]HCl[/tex] solution is 7.1 M
[tex]Molarity=\frac{moles}{\text {volume in L}}[/tex]
[tex]7.1=\frac{moles}{0.0847L}[/tex]
[tex]moles=7.1\times 0.0847=0.60[/tex]
Mass of [tex]HCl=moles\times {\text{Molar mass}}=0.60mol\times 36.5g/mol=21.9g[/tex]
To prepare 7.1 M solution , we need to add 21.9 g of HCl to 84.7 ml of water.
To prepare [tex]1.0\times 10^2ml[/tex] 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.
