Answer :
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
[tex]S.G=\frac{D}{d_w}[/tex]
[tex]d_w[/tex] = density of water = 1 g/mL
[tex]D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL[/tex]
Mass of the solution = 588 g + 1692 g = 2280 g
Volume of the solution = V
Volume = [tex]\frac{Mass}{Density}[/tex]
[tex]=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L[/tex]
1 mL = 0.001 L
[tex]Molarity = \frac{n}{V(L)}[/tex]
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 :
[tex]Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L[/tex]