what mass of zinc chloride, zncl2 is required to prepare 5.00l of a 0.450 molar solution
How many moles do you need? That is M x L = moles.
Then moles = grams/molar mass. You know moles and molar mass, solve for grams.
To calculate the mass of zinc chloride (ZnCl2) required to prepare a 0.450 molar solution of 5.00 L volume, you need to use the formula:
Mass = Molar mass × Moles
First, determine the molar mass of zinc chloride (ZnCl2) by adding up the atomic masses of zinc (Zn) and chlorine (Cl).
Atomic mass of Zn = 65.38 g/mol
Atomic mass of Cl = 35.45 g/mol
Molar mass of ZnCl2 = (1 × Atomic mass of Zn) + (2 × Atomic mass of Cl)
= (1 × 65.38) + (2 × 35.45)
= 65.38 + 70.90
= 136.28 g/mol
Next, calculate the amount of moles using the formula:
Moles = Molarity × Volume
Moles = 0.450 mol/L × 5.00 L
= 2.25 mol
Finally, calculate the mass of zinc chloride:
Mass = Molar mass × Moles
= 136.28 g/mol × 2.25 mol
= 306.39 g
Therefore, to prepare a 0.450 molar solution with a volume of 5.00 L, you will need 306.39 grams of zinc chloride (ZnCl2).