Answer :
Answer: 0.10 L
Explanation:
Given: 37 gram of HCl are dissolved in 100 g of solution.
Density of solution= 1.20 g/ml
Volume of solution = [tex]\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.20g/ml}=83.3ml[/tex]
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\tims 1000}{V_s}[/tex]
where,
n = moles of solute =[tex]\frac{\text{Given mass}}{\text{Molar mass}}=\frac{37g}{36.5}=1.01moles[/tex]
[tex]V_s[/tex] = volume of solution in ml = 83.3 ml
[tex]Molarity=\frac{1.01\tims 1000}{83.3}=12.2M[/tex]
[tex]M_1V_1(stock)=M_2V_2(dilute)[/tex]
[tex]12.2\times V_1=0.500\times 2.5L[/tex]
[tex]V_1=0.10L[/tex]
0.10 L of concentrated stock solution in milliliters should you use to make 2.5 L of 0.500 M HCl
The volume of the concentrated stock solution needed to prepare the solution is 102.71 mL
Let the mass of the concentrated HCl solution be 100 g.
thus, the mass of 37.0% HCl will be 37 g
- Next, we shall determine the volume of the concentrated HCl solution
Mass of solution = 100 g
Density of solution = 1.20 g/ mL
Volume of solution = ?
[tex]volume \: = \frac{mass}{density} \\ \\ volume \: = \frac{100}{1.2} \\ \\ [/tex]
Volume of solution = 83.33 mL
- Next, we shall determine the number of mole in 37 g of HCl. This can be obtained as follow:
Mass of HCl = 37 g
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
mole of HCl =?
[tex]mole \: = \frac{mass}{molar \: mass} \\ \\ mole \: = \frac{37}{36.5} \\ \\ [/tex]
Mole of HCl = 1.014 mole
- Next, we shall determine the molarity of the concentrated HCl solution.
Mole of HCl = 1.014 mole
Volume of solution = 83.33 mL = 83.33 / 1000 = 0.08333 L
Molarity of HCl = ?
[tex]molarity \: = \frac{mole}{volume} \\ \\ molarity \: = \frac{1.014}{0.08333} [/tex]
Molarity of HCl = 12.17 M
- Finally, we shall determine the volume of stock solution needed.
Molarity of stock solution (M₁) = 12.17 M
Volume of diluted solution (V₂) = 2.5 L = 2.5 × 1000 = 2500 mL
Molarity of diluted solution (M₂) = 0.5 M
Volume of stock solution needed (V₁) = ?
M₁V₁ = M₂V₂
12.17 × V₁ = 0.5 × 2500
12.17 × V₁ = 1250
Divide both side by 12.17
V₁ = 1250 / 12.17
V₁ = 102.71 mL
Therefore, the volume of the concentrated stock solution needed to prepare the solution is 102.71 mL
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