Question

27.molarity of 29% w/W H2so4 solution whose density is 1.22 g/ml

Answer 1

• It is given that w/W% of sulphuric acid is 29% which means 29 g of sulphuric acid is dissolved in 100 g of solution
• Density of sulphuric acid = 1.22 g/ml
• Hence, the equation for molarity$Molarity=\frac{no.ofmoles}{volumeinLitre}$

• The number of moles$=\frac{givenmass}{molarmass}=\frac{29}{98}moles$ since the molar mass of sulphuric acid is 98 g

• The volume of sulphuric can be calculated by using the given density= 1.22 g/ml and mass=100 g:

  $density=\frac{mass}{volume}$

  $\therefore volume=\frac{mass}{density}=\frac{100}{1.22}mL$

  $volumeinlitres=\frac{100}{1.22\times 1000}$

• Therefore,

  $\text{Molarity}=\frac{n}{v}=\frac{\frac{29}{98}}{\frac{100}{1.2\times 1000}}$

  $=\frac{29\times 1.2\times 1000}{98\times 100}$

  $=3.61M$

Hence, the molarity of 29% w/W of sulphuric acid, with a density of 1.22 g/ml is 3.61 M