Question

6.8gm of H2O2 is present in 500ml of an aqueous solution.when 10ml of this solution is decomposed what is the stp volume of oxygen obtained​

Answer 1

44.8 ml of oxygen is obtained at STP .

It has given that 6.8 g of H₂O₂ is present in 500ml of an aqueous solution. when 10ml of this is decomposed,

we have to find the volume of oxygen obtained at STP.

here molecular weight of H₂O₂ = 2 + 32 = 34g/mol

given weight = 6.8g

so, the number of moles of hydrogen peroxide = 6.8/34 = 0.2 mol

molarity = number of moles of solute/volume of solution in L

= 0.2 mol/(500/1000 L)

= 0.4 mol/L

now number of moles of hydrogen peroxide in 10ml of solution = Molarity × volume in L

= 0.4 mol/L × 10/1000

= 0.004 mol

now dissociation reaction of hydrocarbon peroxide,

2H₂O₂ ⇔2H₂O + O₂

here it is clear that 2 moles of hydrogen peroxide dissociate into 2 moles of water and one mole of oxygen.

so, 0.004 mol of hydrogen peroxide forms 0.004/2 = 0.002 mol of oxygen.

now volume of oxygen at STP = 22.4 L × 0.002

= 0.0448 L = 44.8 ml