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The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is _____mol/L. (Molecular weight of HNO3 = 63)
(Answer up to the nearest integer)
(Answer up to the nearest integer)
Correct answer is '14'. Can you explain this answer?
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Most Upvoted Answer
The molarity of HNO3in a sample which has density 1.4 g/mL and mass pe...
63% W/W HNO3 solution having density 1.4 g/mL, i.e. 100 g solution, has 63 g of HNO3.
Volume of 100 g of solution = 100/1.4 mL
Volume of 100 g of solution = 100/1.4 mL
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The molarity of HNO3in a sample which has density 1.4 g/mL and mass pe...
Given data:
Density of HNO3 = 1.4 g/mL
Mass percentage of HNO3 = 63%
Molecular weight of HNO3 = 63 g/mol
Calculating the mass of HNO3 in the sample:
Let's assume we have 100 g of the sample.
Mass of HNO3 in the sample = 63% of 100 g = 63 g
Calculating the volume of the sample:
Using the density formula: Density = Mass/Volume
Volume = Mass/Density = 100 g / 1.4 g/mL ≈ 71.43 mL
Calculating the molarity of HNO3:
Now, we need to convert the volume from mL to L: 71.43 mL = 0.07143 L
Molarity = (Mass of solute in grams / Molecular weight) / Volume in liters
Molarity = (63 g / 63 g/mol) / 0.07143 L
Molarity = 1 mol / 0.07143 L ≈ 14 mol/L
Therefore, the molarity of HNO3 in the given sample is approximately 14 mol/L.
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The molarity of HNO3in a sample which has density 1.4 g/mL and mass percentage of 63% is _____mol/L. (Molecular weight of HNO3= 63)(Answer up to the nearest integer)Correct answer is '14'. Can you explain this answer?
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