Question

Find molality of 49% W/W H2SO4

Answer 1

Answer:

Your question is deficient of one data and that is density of solution!

I will assume it to be d g/cm3

Now 49%w/v means 49 gm of H2SO4 in every 100 ml solution.

Taking density of solution to be d g/cm3, total weight of solution = 100d gm

Therefore, weight of water = (100d-49) gm

Moles of H2SO4 = 49/98=0.5

Moles of water= (100d-49)/18

Mole fraction of H2SO4 =

0.5/(0.5+(100d-18)/18)

Now you need to put the value of density to get the exact answer.

You may assume the density of solution to be 1g/cm3 if it is not mentioned in the original question. But that would make the answer approximate.

Hope this helps!

Answer 2

Answer:

m=9.8

Explanation:

49% w/w H2SO4 - 49g h2so4 in 100g solution,mass of solvent=(mass of solution - mass of solute)—>(100–49)=51g of solvent.

Moles of Solute =49/98–>0.5moles of h2so4

Now,51g of solvent has —> 0.5 moles of h2so4

1g of solvent—>0.5/51

1kg of solvent—>(0.5×1000)/51==>9.8

m=moles of solute/1kg of solvent(no. of moles of solute per kg of solvent)

Hope this Helps❤❤✌✌