Calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid.
1 answer:
The question is incomplete,the complete question :
Calculate the molality of a 10.0% (by mass) aqueous solution of hydrochloric acid:
a) 0.274 m
b) 2.74 m
c) 3.05 m
d) 4.33 m
e) the density of the solution is needed to solve the problem
Answer:
The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.
Explanation:
10.0% (by mass) aqueous solution of hydrochloric acid.
10 grams of HCl is present in 100 g of solution.
Mass of HCl = 10 g
Mass of solution = 100 g
Mass of solution = Mass of solute + Mass of water
Mass of water = 100 g - 10 g = 90 g
Moles of HCl =
Mass of water in kilograms = 0.090 kg
Molality =
The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.
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Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:
1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows
10 ml 17.50 ml
(x) M 0.200 M
Molarity =
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
=
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration =
Molar Concentration =
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M