Question

Calculate the molality of 1 litre solution of H2SO4 which is 23% w/v. ​

Answer 1

Answer:

The molality of a 1L solution with x% H2SO4 is 9. The weight of solvent is 910 grams.....

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Answer 2

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missing point :- ( density of soln = 1.2 g / cm³)

23 % w / v means 23 gram of H2SO4 in 100 ml solution

Then ,230 g of H2SO4 is present in 1000 ml solution

mole of H2SO4

= mass of H2SO4 / molar mass of H2SO4

= 230/ 98

= 2.34

mole of H2SO4 = 2.34 moles

density of soln = 1.2 g / cm³

vol of soln = 1000 ml

mass of soln = density of soln × vol of soln

mass of soln = 1.2 ×1000

mass of soln = 1200g

mass of solvent = mass of soln - mass of solute

mass of solvent = 1200- 230

mass of solvent = 970 g

we know that ,

molality = moles of solute / mass of solvent ( kg )

molality = 2.34 /0. 970

molality = 2.41 m