To prepare one liter of 2.5 M solution of HCl, what is the volume of concentrated HCl (37% p/p, d = 1.19 g/mL) to be used?
First determine the M of the concentrated (the 37 %) HCl solution. That is
1000 x 1.19 x 0.37 x (1/36.45) = approx 12 M but you need to do it more exactly.
Then use
mL1 x M1 = mL1 x M2
mL1 x 12M = 1000 mL x 2.5M
Solve for mL1 which is mL of the 37% stuff.
student
ens/047/11
To calculate the volume of concentrated HCl needed to prepare a 2.5 M solution in one liter, we can use the formula:
Molarity (M) = (moles of solute) / (volume of solvent in liters).
First, let's calculate the moles of HCl needed. Since we want to prepare a 2.5 M solution, we can use the equation:
Moles of solute = Molarity × Volume of solvent (in liters).
Moles of solute = 2.5 mol/L × 1 L = 2.5 moles.
Now we need to find the mass of HCl needed using the molar mass of HCl. The molar mass of HCl (hydrochloric acid) is 36.461 g/mol (1 mole of HCl contains 1 atom of hydrogen with a molar mass of 1.0079 g/mol and 1 atom of chlorine with a molar mass of 35.453 g/mol).
Mass of HCl = Moles of HCl × Molar mass of HCl.
Mass of HCl = 2.5 moles × 36.461 g/mol = 91.1525 grams.
Next, we can use the density of the concentrated HCl solution to calculate the volume of concentrated HCl needed. The density of the solution is given as 1.19 g/mL, which means that 1 mL of the solution weighs 1.19 grams.
Volume of concentrated HCl = Mass of HCl / Density of HCl solution.
Volume of concentrated HCl = 91.1525 g / 1.19 g/mL = 76.5 mL.
Therefore, to prepare one liter of a 2.5 M solution of HCl, you would need 76.5 mL of concentrated HCl (37% p/p, d = 1.19 g/mL).