Question

Part 5. Molality
1. A solution of sulfuric acid was made by dissolving 20.0 g of 98% sulfuric acid in
250 ml water. Find the molality of the solution.
2. Find the molality of a solution resulting from the addition of 100 ml of 20% sodium
chloride to 150 ml of 5% NaCl solution. Assume density is 1.1.

PLease answer this question kindly...Don't comment with false answer huhuhu.. thank u for understanding (feeling grateful)

Answer 1

Explanation:

molality =moles of solute/mass of solvent in kg

98% sulphuric acid means in 100 g sulphuric acid source exact amount of sulphuric acid, H2SO4 present 98 g.

So in 20 g, exact amount of H2SO4 present

= (98×20) \100

= 19.6 g H2SO4 present

moles of H2SO4=19.6/98=0.2 moles

mass of water =volume of water(density is 1)

molality of H2SO4=0.2/0.25=0.8