We dissolve 10.0 g of each of the following in 800 g of water. In which solution would the molality of solute be highest? 1. All
1 answer:
You might be interested in
The volume of H₂ evolved at NTP=0.336 L
<h3>Further explanation</h3>Reaction
Decomposition of NH₃
2NH₃ ⇒ N₂ + 3H₂
conservation mass : mass reactants=mass product
0.28 NH₃= 0.25 N₂ + 0.03 H₂
2 g H₂ = 22.4 L
so for 0.03 g :
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