Question

Calculate the molarity of zn2 in 150.0 ml of aqueous solution that contains 5.07 g of zinc chloride.

a. 0.248 mol/l
b. 248 mol/l
c. 2.48 mol/l
d. 24.8 mol/l

Answer 1

The molarity of Zn2+ in a 150.0 mL solution containing 5.07 g of zinc chloride is calculated to be approximately 0.248 M (option a).

Step-by-step explanation:

To calculate the molarity of Zn2+ in a 150.0 mL aqueous solution containing 5.07 g of zinc chloride (ZnCl2), use the following steps:

1. Calculate the moles of ZnCl2 using its molar mass.
   
   The molar mass of ZnCl2 is approximately 136.3 g/mol (65.38 g/mol for Zn + 2 × 35.45 g/mol for Cl). So:
2. Moles of ZnCl2 = mass (g) / molar mass (g/mol)
   
   Moles of ZnCl2 = 5.07 g / 136.3 g/mol
   
   Moles of ZnCl2 ≈ 0.0372 mol
3. Since each mole of ZnCl2 produces one mole of Zn2+ ions, the moles of Zn2+ will also be 0.0372 mol.
4. Convert the volume of the solution from mL to L.
   
   150.0 mL = 0.150 L
5. Calculate the molarity (M) of Zn2+ ions:
   
   Molarity (M) = moles of solute (mol) / volume of solution (L)
   
   Molarity (M) = 0.0372 mol / 0.150 L
   
   Molarity (M) ≈ 0.248 M

The molarity of Zn2+ in the solution is 0.248 mol/L.