Question

Commercially available sulphuric acid contains 93% acid by mass and has a density of 1.84 gm/ cm^ 3 . calculate the molarity of the solution .

Answer 1

Final Answer : 17.46 M
Formula used:
1) No. of moles = Given mass /Molar Mass
2) Density = Mass /Volume
3) Molarity =No. of moles of solute /Volume of solution

Steps:
1) Let the total mass of solution be 100g .
Then, Mass of Sulphuric acid = 93g
Mass of water = 7g
Molar mass of Sulphuric acid = 98g.
Molar Mass of water = 18g

No. of moles of sulphuric acid =93/94

2) Density of pure sulphuric solution = 1.84 g/mL
Volume of sulphuric acid = 100/1.84 mL=54.34 mL


3) Molarity =
$= \frac{ \frac{93}{98} }{( 54.34 ) \times {10}^{ - 3} } \\ = \frac{ \frac{93}{98} }{50.54} \times {10}^{3} \\ = 17.46M molar \:$

Molarity of Sulphuric Acid in its solution = 17.46M

Answer 2

% of the sulfuric acid in the solution = 93 % 

This means that 93 g of the sulphuric acid is present in the 100 g of the solution. 

Molar Mass = 98 g/mole. 

∵ No. of moles = Mass/Molar mass 
∴ No. of moles = 93/98
  = 0.949 moles. 

Also, Density of the solution = 1.84 g/cm³.

∵ Density = Mass/Volume. 
∴ Volume = 100/1.84
    = 54.35 cm³
    = 0.543 L.


Now, Using the Formula, 

  Molarity = No. of moles of solute/Volume of the solutions in liter. 
     = 0.95/0.0543
     = 17.5 M.


Hence, the molarity of the solution is 17.5 M. 


Hope it helps.