a bottle of aqueous H2O2 solution which is labeled '28V' H2O2 and density of the solution is 1.25 gm/ ml. Choose the correct option a) Molality of H2O2 is 2. b) Molarity of H2O2 is 5 c) Molality of H2O2 is 2.15 d) none of these
This means that 1 volume of H2O2 will give off 28 volumes of O2 at NTP, 20oC (298.15K) and 1 atm.
So, 1 L of solution can produce 28 L of oxygen under these conditions.
Let's assume oxygen is a perfect gas, then from
PV=nRT where P = 1 atm and V = 28 L, we can calculate n, the number of moles of O2
n = PV/RT = (1)(28) / (0.0821 X 298.15)
n = 1.144 moles of O2
Hydrogen peroxide decomposes as 2H2O2 → 2H2O + O2
Since twice the no of moles are required to produce one mole of oxygen so to generate 1.144 moles of O2 in 1 L of the solution, there must be 2.28 moles of H2O2 present in 1 L.
So molarity of H2O2 is 2.28
Density = 1.25 g/ml
So mass of 1000ml solution = 1250 g = 1.25 kg
So molality = 2.28/1.25 = 1.824 m which is none of the given values. So answer is (d).
So, 1 L of solution can produce 28 L of oxygen under these conditions.
Let's assume oxygen is a perfect gas, then from
PV=nRT where P = 1 atm and V = 28 L, we can calculate n, the number of moles of O2
n = PV/RT = (1)(28) / (0.0821 X 298.15)
n = 1.144 moles of O2
Hydrogen peroxide decomposes as 2H2O2 → 2H2O + O2
Since twice the no of moles are required to produce one mole of oxygen so to generate 1.144 moles of O2 in 1 L of the solution, there must be 2.28 moles of H2O2 present in 1 L.
So molarity of H2O2 is 2.28
Density = 1.25 g/ml
So mass of 1000ml solution = 1250 g = 1.25 kg
So molality = 2.28/1.25 = 1.824 m which is none of the given values. So answer is (d).