Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL-1?
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.
Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol - 1
Then, number of moles of HNO3 = 68 / 63 mol
= 1.08 mol
Also density = 1.504g/mL-1 (given)
Therefore from the formula density = mass / volume, we get
Volume of solution = 1000/1.504 = 66.49 mL
Therefore molarity of nitric acid = (1.08/66.49) x 1000 = 16.24 M
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Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate of formation of D/mol L - 1 min - 1 |
I | 0.1 | 0.1 |
6.0 × 10 - 3 |
II | 0.3 | 0.2 |
7.2 × 10 - 2 |
III | 0.3 | 0.4 |
2.88 × 10 - 1 |
IV | 0.4 | 0.1 |
2.40 × 10 - 2 |
Determine the rate law and the rate constant for the reaction.
In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
A/ mol L - 1 |
0.20 | 0.20 | 0.40 |
B/ mol L - 1 |
0.30 | 0.10 | 0.05 |
r0/ mol L - 1 s - 1 |
5.07 × 10 - 5 |
5.07 × 10 - 5 |
1.43 × 10 - 4 |
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Thanks
100g solution not 1000
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Ise wayer dencity 100 not 1000
Volume should be 100 not 1000ð
The volume of the solution should be 100g not 1000g
Volume of the solution should 100gnot 1000g
Vol=45.21mL Molarity=23.88M
Vol=45.21mL Molarity=23.88M