The concentration of a hypothetical molecule, A5B6C2, is 943 mg/dL. The molar masses of the hypothetical...
The concentration of a hypothetical molecule, A5B6C2, is 943 mg/dL. The molar masses of the hypothetical elements are in the table below. What is the molarity of this molecule in the solution?
| Element | Molar Mass (grams/mole) |
| A | 10.852 |
| B | 1.0797 |
| C | 15.35 |
Homework Answers
Molecular mass of A5B6C2 = (10.852 x 5) + (1.0797 x 6) + (15.35 x 2) = 54.26 + 6.4782 + 30.7 = 91.4382 g
It is given in the question that concentration is 943 mg/dl which means that 943 mg of solute in 1 dl of water.
No. of moles of solute = Given Mass / Molecular Mass = 943 mg / 91.4382 g
1 mg = 0.001 g
No. of moles of solute = Given Mass / Molecular Mass = 943 X 0.001 g / 91.4382 g = 0.0103 moles
Given volume = 1 dl = 0.1 litre
Molarity = Number of moles of solute / Volume of solution in litre = 0.0103 / 0.1 mol/l = 0.103 mol/l
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