Question

Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)

Answer 1

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

$S.G=\frac{D}{d_w}$

$d_w$ = density of water = 1 g/mL

$D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL$

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = $\frac{Mass}{Density}$

$=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L$

1 mL = 0.001 L

$Molarity = \frac{n}{V(L)}$

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

$Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L$