A commercially available sample of sulphuric acid is 15% H2SO4 by mass (density=1.10g/ml) Calculate the molarity ,normality and molality.

Answer 1

15% w/w means 15 g H2SO4/100 g solution or
15 g H2SO4/(15 + 85)g soln.
mols H2SO4 = 15/98 = about 0.15 but you do it more accurately.
Use the density to convert 100 g solution to volume, change that to L, then
M = mols/L.

That should get you started.

Answer 2

3.3646N

Answer 3

15% H2SO4 by weight means 15g of
H2SO4 dissolved in 85g of water
Density=mass/volume
volume=100/1.10=90.90ml (V)

MOLARITY=WB/MB*1000/V
15/98*1000/90.90=1.68M

Answer 4

15% solution of H2So4 mean that 15gm of H2So4 is present in 100gm of solution or 85gm of solvent.
Molarity =Mass of the solution in gm×1000 /Molar mass ×Volume of the solution in ml
=15×1000/98×90.90
=1.68M .

Answer 5

15% solution of H2So4 mean that 15gm of H2So4 is present in 100gm of solution or 85gm of solvent.
Molarity =Mass of the solute ×1000 /Molar mass ×Volume of the solution in ml
=15 ×1000 /98×90.90
=1.68M
Normality =Mass of the solution in gm ×1000 /Volume of the solution in ml ×Equivalent mass
=100×1000/90.90×46
=3.36N
Molality =Mass of the solute ×1000/Mass of the solvent in gm ×Molar mass
=15×1000/85×98
=1.8m .
- End -
I hope you want my answer. Thank you .lam Sarju Elangbam from Khangabok, Manipur.

Answer 6

Not good

Answer 7

Thank you so much sir

Answer 8

Thanx

Answer 9

bababooey

Answer 10

21n