commercial HCL IS SOLD AT 37% WITH SPECIFIC GRAVITY OF 1.18.HOW WILL YOU PREPARE 5OML FOR 100STUDENTS AT THE CONCENTRATION OF 0.25M
To prepare 50 mL of hydrochloric acid (HCl) solution for 100 students at a concentration of 0.25 M, we need to calculate the amount of concentrated HCl and water required.
First, let's determine the volume of concentrated HCl needed to make the 0.25 M solution.
1. Start with the desired concentration of HCl: 0.25 M.
2. The concentration of the commercial HCl is given as 37%, which means it contains 37 g of HCl per 100 mL of solution.
3. Convert the desired concentration to moles per liter (M). Since 1 M equals 1 mol/L, the desired concentration will be 0.25 mol/L.
4. Calculate the moles of HCl required: Moles = concentration (mol/L) * volume (L)
Moles = 0.25 mol/L * 0.05 L (50 mL converted to liters)
Moles = 0.0125 mol
Now, let's determine the volume of the concentrated HCl needed.
5. The known concentration of the commercial HCl solution is 37% or 37 g/100 mL.
6. Convert the grams of HCl to moles using the molar mass of HCl (36.46 g/mol):
Moles = mass in grams / molar mass
Moles = 37 g * (1 mol/36.46 g)
Moles = 1.014 mol
7. Use the moles of concentrated HCl and its molar mass to calculate the required volume:
Volume = moles / concentration
Volume = 0.0125 mol / 1.014 mol/L
Volume ≈ 0.012 L or 12 mL
Now, let's determine the volume of water required to dilute the concentrated HCl to a total volume of 50 mL:
8. Subtract the volume of concentrated HCl from the total volume:
Volume of water = 50 mL - 12 mL
Volume of water = 38 mL
To prepare the 0.25 M HCl solution for 100 students:
9. Multiply the volume of concentrated HCl and water by the number of students:
Volume of concentrated HCl = 12 mL * 100 students = 1200 mL (or 1.2 L)
Volume of water = 38 mL * 100 students = 3800 mL (or 3.8 L)
Finally, mix 1.2 L of concentrated HCl with 3.8 L of water to obtain a total volume of 5 L of 0.25 M HCl solution.