1. Find the molarity and molality of a 15% solution of H2SO4 when it's density is 1.10 glcm3 and molar mass = 98 amu .
2. calculate the mole fraction of ethanol and water in a
sample of rectified spirit which contains 46 % ethanol by mass?
3. calculate the % composition in terms of mass of a solution
obtained by mixing 300g of a 25% and 400 g of 40%solution by mass.
4.One litre of sea wator Weight 1030g and contains about 6×10^ -3 g et dissolves O2. Calculate the concentration of dissolved oxygen in ppm?
5.The density of 85% phosphoric acid is 1.70g/cm3.
What is the volume of a solution that contains 17g
of phosphoric acid?
answer these questions....
Answers
Answer:
(1)
15% H₂SO₄ solution means 15 g of H₂SO₄ is present in 100 g of Water.
No. of moles of H₂SO₄ = 15 g / 98 g/mol = 0.153 mol
Weight of solvent = 100g - 15g = 85g = 85 × 10^-3 kg
Molality = Moles of solute / Weight of Solvent in kg
= 0.153 / (85 × 10^-3)
= 1.8 m
Molality of given solution is 1.8 m ……[1]
Volume of solution = Mass of solution / Density
= 100 g / 1.1 g/cm^3
= 90.90 cm^3
= 0.090 Litres
Molarity = Moles of solute / Volume of solution in litres
= 0.153 / 0.090
≈ 1.7 M
Molarity of given solution is 1.7 M ……[2]
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Explanation:
(2) second question ki photo upper Hai
Explanation: Rectified spirit which contains 46% ethanol by mass. In 100 g spirit has 46 grams of ethanol and 54 grams of water. Mole fraction of ethanol is 0.25. (3) Mass of solute (1) = 300g × 25/100
= 75g
Mass of solute (2) = 400g ×40/100
= 160g
Total masss of solute = 75+160
= 235g
Total mass of solution = 300 + 400
= 700g
% composition by mass of solute = mass of solute / total mass of solution ×100
= 235/700 ×100
= 33.57% of solute
% composition by mass of solvent = 100 - 33.57
= 66.43%
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Answer:
above is the correct answer