How does the periodic table make it easier for the scientist to work with and study elements
2 answers:
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Q1)
This is a strong acid- strong base base reaction, balanced equation for the reaction is as follows;
LiOH + HCl ---> LiCl + H₂O
stoichiometry of acid to base is 1:1
volume of HCl used up - 13.60 - 1.25 = 12.35 mL
volume of LiOH used up - 11.20 - 2.65 = 8.55 mL
molarity of LiOH - 0.140 M
The number of LiOH moles reacted -
= 0.001197 mol
according to stoichiometry, number of LiOH moles = number of HCl moles
Therefore number of HCl moles reacted - 0.001197 mol
The number of HCl moles in 12.35 mL - 0.001197 mol
Then number of HCl moles in 1000 mL -
Molarity of HCl - 0.0969 M
Q2)
Volume of HCl used - 12.35 mL
Volume of LiOH used - 8.55 mL
Molarity of HCl - 0.140 M
In 1 L solution of HCl there are 0.140 mol of HCl
Therefore number of HCl moles in 12.35 mL -
Number of HCl moles reacted - 0.001729 mol
since molar ratio of acid to base is 1:1
the number of LiOH moles that reacted - 0.001729 mol
Therefore number of moles in 8.55 mL - 0.001729
Then number of LiOH moles in 1000 mL -
molarity of LiOH - 0.202 M
This is a strong acid- strong base base reaction, balanced equation for the reaction is as follows;
LiOH + HCl ---> LiCl + H₂O
stoichiometry of acid to base is 1:1
volume of HCl used up - 13.60 - 1.25 = 12.35 mL
volume of LiOH used up - 11.20 - 2.65 = 8.55 mL
molarity of LiOH - 0.140 M
The number of LiOH moles reacted -
according to stoichiometry, number of LiOH moles = number of HCl moles
Therefore number of HCl moles reacted - 0.001197 mol
The number of HCl moles in 12.35 mL - 0.001197 mol
Then number of HCl moles in 1000 mL -
Molarity of HCl - 0.0969 M
Q2)
Volume of HCl used - 12.35 mL
Volume of LiOH used - 8.55 mL
Molarity of HCl - 0.140 M
In 1 L solution of HCl there are 0.140 mol of HCl
Therefore number of HCl moles in 12.35 mL -
Number of HCl moles reacted - 0.001729 mol
since molar ratio of acid to base is 1:1
the number of LiOH moles that reacted - 0.001729 mol
Therefore number of moles in 8.55 mL - 0.001729
Then number of LiOH moles in 1000 mL -
molarity of LiOH - 0.202 M
Answer:- Actual molarity of the original sulfuric acid solution is 17.0M.
Solution:- Barium chloride reacts with sulfuric acid to make a precipitate of barium sulfate. The balanced equation is written as:
From this equation there is 1:1 mol ratio between barium sulfate and sulfuric acid. So, if excess of barium chloride is added to sulfuric acid then the moles of sulfuric acid would be equivalent to the moles of barium sulfate. Moles of barium sulfate could be calculated from the mass of it's dry precipitate.
Molar mass of barium sulfate is 233.4 grams per mol. The calculations for the moles of sulfuric acid are given below:
=
From given information, 10.00 mL of final acid solution were taken to react with excess of barium chloride. It means 0.00170 moles of sulfuric acid are present in 10.0 mL of final acid solution. We could calculate the actual molarity of the final solution from here as:
10.0 mL = 0.0100 L
= 0.170M
Now we would use the dilution equation to calculate the actual molarity of the original sulfuric acid solution. The molarity equation is:
From given information, 10.0 mL of original acid solution were taken in a 100 mL flask and water was added up to the mark. It means the 10 fold dilution is done. 10 fold dilution means the molarity becomes one tenth of it's original value. Let's do the calculations in reverse way as we have calculated the molarity of the final solution.
let's say the molarity after first dilution is Y. the volume is taken as 10.0 mL. Final volume is 100 mL and the molarity is 0.170M. Let's plug in the values in the equation:
Y(10.0mL) = 0.170M(100mL)
X = 17.0M
Hence, the actual molarity of sulfuric acid solution is 17.0M.