The molality of calcium chloride (CaCl2) in an aqueous solution is 2.46 m. What is mole fraction of the solute?
1 answer:
Given that CaCl2 = 2.46 m
Therefore 2.46 mole of CaCl2 present in 1.00 g of solution
Mass of 2.46 mole CaCl2= number of mole * molar mass
= 2.46 mole * 110.98 g/mol
= 273.01 g
Mass of water or solvent = mass of solution – mass of solute
= 1000 g -273.01 g
= 726.99 g
Mole of water = amount in g / molar mass
= 726.99 g/18.02 g/ mole
= 40.34 moles
Mole fraction of CaCl2 = number of mole of CaCl2 / total moles
= 2.46 /2.46+40.34
= 2.46/42.8
= 0.057
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The pH of the solution is 2.54.
Explanation:
pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.
The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.
So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the
[HB] is the concentration of base.
Then
So the pH of the solution is 2.54.