Calculate the molarity of Cl− anions in the chemist's solution.Be sure your answer is rounded to the correct number of significant digits.

Question

A chemist prepares a solution of vanadium

III

 chloride 

VCl3

 by measuring out 9.6mg

 of VCl3

 into a 50.mL

 volumetric flask and filling to the mark with distilled water.Calculate the molarity of 

Cl−

 anions in the chemist's solution.Be sure your answer is rounded to the correct number of significant digits.

Answer 1

9.6 mg has 2 sig.figs. and 50. ml has 2 sig, figs. Answer will have 2 sig. figs.

VCl₃ ==> V³⁺ + 3Cl^-

molar mass VCl₃ = 50.9 + (3*35.5) = 50.9 + 106.5 = 157.4 g / mol

moles VCl₃ used = 9.6 mg x 1 g / 1000 mg x 1 mol / 157.4 g = 6.099x10^-5 moles VCl₃

Since each mole of VCl₃ produced 3 moles of Cl^-, we will multiply this value by 3 to get [Cl^-]

6.099x10^-5 x 3 = 1.8x10^-4 moles

To convert to molarity, we will divide by the number of liters of solution

50 ml x 1 L / 1000 ml = 0.05 L

[Cl-] = 1.8x10^-4 mols / 0.05 L = 3.6x10^-3 M