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Answer:
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Explanation:
Mass percentage of oxalic acid = 5.50%
This means that in 100 grams of solution there are 5.50 grams of oxalic acid.
Mass of solution , m = 100
Volume of the solution = V
Density of the solution = d = 1.024 g/mL
V = 97.66 mL = 0.09766 L
(1 mL = 0.001 L)
Moles of oxalic acid =
The molarity of the solution :
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M