Question

the molality of a 1L sol. with x% h2so4 is 9. The weight of solvent present in the sol. is 910 gm The value of x is?

Answer 1

HELLO THERE!

We know, that:

Molality = Moles of solute / Mass of solvent (in kilograms)

here, Molality is 9 m, mass of solvent is 910 gm.

Put these values, we get:

Moles of H₂SO₄ = Molality x Mass of solvent

$= 9\times \frac{910}{1000} = 8.19 mol$

8.19 moles of sulphuric acid = 8.19 x 98 grams = 802.62 (since molecular mass of sulphuric acid is 98).

So, 802.62 grams of  H₂SO₄ is present in the solution.

Total mass of the solution = 910 g + 802.62 g = 1712.62 grams.

Now,

[tex] H_{2}SO_{4} (\frac{m}{m}) = \frac{802.62}{1712.62}\times 100 \\\\= 46.86 [/tex]

Since, Percentage (m/m) = Mass of solute / Mass of solution x 100%

Hence, 46.86% Sulphuric acid is present in the solution, or

x = 46.86%

THANKS!