Question

the given sample of sulphuric acid was found to have mole fraction of h2s4 as0.15 .calculate the molarity of the solution.

Answer 1

I think Molality is asked since density of solution is not provided .

Final Answer :9.8 Molality

Steps:
1) Let the total no. of moles of H2SO4 and water be 1 mole.
=> Moles of H2SO4 = 0.15
=> Moles of H2O = 0.85

2) Molality = No. of moles of solute / Weight of solvent
= No. of moles of H2SO4 / Weight of water in it (in Kg)

$= \frac{0.15}{0.85 \times mm(water)} \\ = \frac{0.15}{0.85 \times 18 \times {10}^{ - 3} } \\ = 9.8 \: units$
Hence, Molality of solution is 9.8 Molal.

Answer 2

Answer:

Explanation:

H2so4 = 0.15 g

H2O = 0.85 g

Molarity = November. Of moles of solute / volume of solution in liter.

So, Molarity = 0.15 / 0.85×18×10 -3

=9.8 unit.

Molarity =9.8 unit.