Question

What is a molarity of a 13% solution
(by weight) of H2 so4? Its density is
1.09g/ml.​

Answer 1

Answer

1.44 M

Explanation

Let the volume of solution be x ml.

Then, it is given that the weight by weight percentage of solution is 13%

→ weight of H2SO4/weight of water × 100 = 13

We know that density of solution is 1.09 g/ml

→ x ml of solution would weigh 1.09x g.

Then,

weight of H2SO4 = 13 × 1.09 x/100

→ weight of H2SO4 = 14.17x/100

Now, moles of H2SO4 = given weight/molar mass

→ 14.17x/100 × 1/98

→ 14.17x/9800

Now, molarity = mole/volume in litre.

→ x ml = x/1000 litres.

So, molarity = 14.17x/9800 × 1000/x

→ 141.7/98

→ 1.44 M (approx)

Answer 2

Answer:

hope This Answer Help You

Explanation:

Molarity: moles of solute in 1litre of solution

Density of solution=1.09g/ml

This means 1.09g of solution is present in 0.001l

1.09 in 0.001l

Mass of the solution in it's 1l=1.09×1000g

Mass of sulphuric acid=13% mass of the solution

Thus,mass of the sulphuric acid present in 1l of solution=0.13×1.09×1000=13×10.9g

Moles of sulphuric acid in 1l of solution=mass/molar mass=13×10.9/98=1.446

So, molarity=1.446m

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