Given data : The molarity of aqueous solution of unknown compound = 2.1 M The de
ID: 700481 • Letter: G
Question
Given data:
The molarity of aqueous solution of unknown compound = 2.1 M
The density of solution = 0.832 g/cm3
1 cm3 = 1 mL
Therefore, the density of solution = 0.832 g/mL
The molar mass of the unknown compound = 45 g/mol
The molality of the solution of the unknown compound =?
Formula: Molarity = {no. of moles / volume of solution (mL)} * 1000
Molality = {no. of moles / Weight of solution (g)} * 1000
Molarity/molality = {no. of moles / volume of solution (mL)} * 1000 / {no. of moles / Weight of solution (g)} * 1000
i.e. 2.1 / molality = Weight of solution (g) / volume of solution (mL)
The density of solution = weight / volume of solution
Therefore, 2.1/molality = density of solution (in g/mL)
i.e. 2.1 / molality = 0.832
i.e. molality = 2.1 / 0.832
Therefore, the molality of solution = 2.524 mol/Kg
Explanation / Answer
(8 Points) An aqueous solution of an unknown compound is 2.10 M and the density of the is in the solution? The molar mass of the unknown compound is 45.0 g/mo. e solution is 0.832 g/cm. What is the molality of this compound
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.