bottle of hcl which is 40 mass%and has density of 1.20g /ml. calculate ml to be taken from this bottle to prepare 500 ml solution of 0.2 M
First, determine the molarity of the solution of HCl and
Second, use the dilution formula.
1.20 g/mL x 1000 mL x 0.40 x (1 mol/36.5g) = 13.15 which I would round to 13.2 M
Then c1v1 = c2v2
13.2*v1 = 0.2M x 500 mL
Solve for v1.
You can do the M of the initial HCl solution another way.
40% by mass means 40g HCl/100 g solution.
40 g is how many mols? That's mols = grams/molar mass = 40/36.5 = 1.096 mols in 100 g solution. What's the volume of the solution (the 100g?). That's mass = volume x density or volume = mass/density = 100g/1.20 = 83.33 mL or 0.08333 L, so M = mols/L = 1.096/0.083333 = 13.15 which rounds to 13.2,
wrong question
To calculate the volume of the 40% HCl solution needed to prepare a 500 mL solution of 0.2 M HCl, you would need to use the formula:
(Desired Molarity * Desired Volume) / Concentration of HCl in the solution
Let's begin by finding the concentration of HCl in the 40% HCl solution:
Concentration = Mass of HCl / Volume of HCl solution
Given that the density of the solution is 1.20 g/mL, we can calculate the mass of HCl in 100 mL (which is equal to 40% of the total volume of the solution):
Mass of HCl = Volume of HCl solution * Density of HCl solution * Concentration of HCl
Mass of HCl = 100 mL * 1.20 g/mL * 0.40
Now, we can calculate the volume of the 40% HCl solution needed to prepare the 500 mL solution of 0.2 M HCl:
Volume of 40% HCl solution = (Desired Molarity * Desired Volume) / Concentration of HCl
Volume of 40% HCl solution = (0.2 M * 500 mL) / (Mass of HCl / Volume of HCl solution)
Finally, substitute the values into the formula to get the result:
Volume of 40% HCl solution = (0.2 M * 500 mL) / ((100 mL * 1.20 g/mL * 0.40) / 100 mL)
Simplifying the expression:
Volume of 40% HCl solution = (0.2 M * 500 mL) / (0.48 g/mL)
Volume of 40% HCl solution = 833.33 mL
Therefore, you would need to take 833.33 mL from the bottle of 40% HCl to prepare a 500 mL solution of 0.2 M HCl.