Convert a CaCL2 standard with a concentration of 8.40 mg/dL to mEq / L.

Question

You added 400 mg of NaHCO3 to 500 mL of diluent, what is the molarity of this solution?

Convert a CaCL2 standard with a concentration of 8.40 mg/dL to mEq / L.

Answer 1

molar mass CaCl₂ = 111 g / mole

For CaCl₂, there are 2 milli equivalents per millimole.

mmoles CaCl₂ = 8.40 mg x 1 mmole / 111 mg = 0.07568 mmoles

Liters of solution = 1 dL x 1 L / 10 dL = 0.1 L

mmol / L = 0.07568 mmols / 0.1 L = 0.7568 mmol / L

mEq / L = 0.7568 mmol / L x 2 mEq / mmol = 1.51 mEq / L (3 sig. figs.)

molar mass NaHCO3 = 84.0 g / mol

400 mg x 1 g / 1000 mg x 1 mol / 84.0 g = 0.004762 moles

500 ml x 1 L / 1000 ml = 0.500 L

molarity = moles / L

Assuming no change in volume when adding the NaHCO3, the molarity will be ...

0.004762 moles / 0.500 L = 0.00952 M = 9.52x10^-3 M (3 sig. figs.)