1. What are these bubbles? d.evaporated water 7. Calculate the concentration of
ID: 1042783 • Letter: 1
Question
1. What are these bubbles?
d.evaporated water
7.
Calculate the concentration of each solution in mass percent.
Part A
147 g KCl in 568 g H2O
Part B
30.3 mg KNO3 in 7.81 g H2O
Part C
8.75 g C2H6O in 74.5 g H2O
15. Describe how you would make 100.0 mL of a 1.00 M NaOH solution from a 17.0 M stock NaOH solution.
Express your answer with the appropriate units.
d.evaporated water
Explanation / Answer
Answer:
1. Bubbles in solution: gases that are coming out of walls of the pot
Like differences of CO2.
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7.
Part A: Mass of KCl = 147.0 g, Mass of water = 568.0 g
Total Mass of (KCl+Water) solution = 147.0 + 568.0 = 715.0 g
% Mass of KCl = (Mass of KCl / Total mass)*100 = (147.0/715.0)*100 = 20.56%
% Mass of water = (Mass of Water / Total mass)*100 = (568.0/715.0)*100 = 79.44%.
Part B: Mass of KNO3 = 30.3 mg, Mass of water = 7.810 g = 7810.0 mg
Total mass of (KNO3+Water) solution = 30.3 + 7810.0 = 7840.3 mg
% Mass of KNO3 = (Mass of KNO3 / Total mass)*100 = (30.3/7840.3)*100 = 0.39%
% Mass of water = (Mass of Water / Total mass)*100 = (7810.3/7840.3)*100 = 99.61%.
Part C: Mass of C2H6O = 8.75 g, Mass of Water = 74.50 g
Total mass = 8.75 + 74.5 = 83.25
% mass of C2H6O = (8.75 / 83.25)*100 = 10.51 %
% mass of Water = (74.50 / 83.25)*100 = 89.49 %
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15)
Let we need solution with,
M1 = 1.00 M & V1 = 100.0
Stock solution has, M2 = 17.0 M V2 = ?
By law of dilution
M2*V2 = M1*V1
17.0 M * V2 = 1.00 M * 100.0 mL
V2 = 1.00 M * 100.0 mL / 17.0 M
V2 = 5.9 mL
5.9 mL of 17.0 M NaOH stock solution taken in a 100.0 mL standard flask.
To this added sufficient amount of DI water up to 100.0 mL mark.
Hence the 100.0 mL of 1.00 M NaOH solution
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