The question is incomplete, the complete question is:
A chemist prepares a solution of vanadium (III) chloride (VCl3) by measuring out 0.40g 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:
0.153M of anions
Explanation:
First we calculate the concentration of the solution. From m/M= CV
m=given mass, M= molar mass, C =concentration of solution, V= volume of solution
Molar mass of compound= 51 + 3(35.5)= 157.5gmol-1
0.4g/157.5gmol-1= C×50/1000
C= 2.54×10-3/0.05= 0.051M
But 1 mole of VCl3 contains 3 moles of anions
Therefore, 0.051M will contain 3×0.051M of anions= 0.153M of anions
Answer:
See explanation below
Explanation:
You are not providing the value of the vanadium chloride and the volume used.
For this problem, I will use 0.30 g of Vanadium chloride and a volumetric flask pf 50 mL to do this. You later, only need to replace your data to get the accurate answer.
First let's write the reaction of solution of Vanadium chloride:
VCl₃ ------> V³⁺ + 3Cl⁻
This means that 1 mole of VCl₃ produces 3 moles of Cl⁻. So, to get the concentration, we need to calculate the moles of VCl₃ and then, calculate the moles of the chlorine:
The molecular mass of VCl₃ is 157.3 g/mol, so the moles are:
n = 0.30 / 157.3 = 0.0019 moles
now if 1 mole VCl₃ -----> 3 moles Cl then:
moles Cl⁻ = 0.0019 * 3 = 0.0057 moles
Finally to get the concentration:
M = n/V
the volume is 50 mL or 0.050 L so the concentration would be:
M = 0.0057 / 0.050
M = 0.114 M
This would be the concentration of the anions
