Calculate the moles of H2SO4 present in 100g of 20m H2SO4 solution.
m= molality
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Answers
Answer:
Hello Dear.
☆ Here is the answer ☆
Given
Specific gravity = 1.3
Since,
Specific Gravity = Density of the solution/Density if the Water.
Now, Density of Water = 1 g/ml.
Density of Water = 1.3 × 1 g/ml.
= 1.3 g/ml.
Density of Sulphuric Acid = 100 g/liter.
It means that 100 g of suphuric acid is present in 1 liter of solution.
Since,
No. Of moles of sulphuric acid = Mass/Molar Mass.
= 100/98
= 1.02 moles.
For Calculating Molarity,
Using the Formula,
Molarity = No. Of moles of solute/Volume of the solution in liter.
= 1.02/1
= 1.02 M.
For Calculating Molality,
Volume of the Solution = 1 liter.
= 1000 ml.
Density of the Solution = 1.3 g/ml.
Therefore,
Mass of the solution = 1.3 × 1000
= 1300 g.
Now, Mass of the Solvent = Mass of the Solution - Mass of the Solute.
Mass of the Solvent = 1300- 100
= 1200 g.
= 1.2 kg.
Using the Formula,
Molality = No. Of moles of Solute/Mass of the Solvent in kg.
= 1.02/1.2
= 0.84 m.
Thus, Molality of the solution is 0.84 m.
☆ ■ Hope it helps. ■ ☆
Answer:
Explanation:
Since,
Specific Gravity = Density of the solution/Density if the Water.
Now, Density of Water = 1 g/ml.
Density of Water = 1.3 × 1 g/ml.
= 1.3 g/ml.
Density of Sulphuric Acid = 100 g/liter.
It means that 100 g of suphuric acid is present in 1 liter of solution.
Since,
No. Of moles of sulphuric acid = Mass/Molar Mass.
= 100/98
= 1.02 moles.
For Calculating Molarity,
Using the Formula,
Molarity = No. Of moles of solute/Volume of the solution in liter.
= 1.02/1
= 1.02 M.
For Calculating Molality,
Volume of the Solution = 1 liter.
= 1000 ml.
Density of the Solution = 1.3 g/ml.
Therefore,
Mass of the solution = 1.3 × 1000
= 1300 g.
Now, Mass of the Solvent = Mass of the Solution - Mass of the Solute.
Mass of the Solvent = 1300- 100
= 1200 g.
= 1.2 kg.
Using the Formula,
Molality = No. Of moles of Solute/Mass of the Solvent in kg.
= 1.02/1.2
= 0.84 m.
Thus, Molality of the solution is 0.84 m.