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Sagot :
The mass percent of HNO₃ in molarity is 15.76 M.
Assumption
Let the mass of the solution be 100 g.
Therefore, the mass of 70% HNO₃ in the solution is 70 g
Determination of the mole of HNO₃
- Mass = 70 g
- Molar mass of HNO₃ = 1 + 14 + (3×16) = 63 g/mol
- Mole of HNO₃ =?
Mole = mass / molar mass
Mole of HNO₃ = 70 / 63
Mole of HNO₃ = 1.11 mole
Determination of the volume of the solution
- Mass of solution = 100 g
- Density of solution = 1.42 g/cm³
- Volume of solution =?
Volume = mass / density
Volume of solution = 100 / 1.42
Volume of solution = 70.42 cm³
Divide by 1000 to express in L
Volume of solution = 70.42 / 1000
Volume of solution = 0.07042 L
Determination of the molarity
- Mole of HNO₃ = 1.11 mole
- Volume of solution = 0.07042 L
- Molarity =?
Molarity = mole / Volume
Molarity = 1.11 / 0.07042
Molarity = 15.76 M
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