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H2O2 in aqueous solution decomposes on warming as: 2H2O2(aq) → 2H2O(l) + O2(g) If 1 mole of gas occupies a volume of 25 L under the conditions of measurement and 200 ml of x M solution of H2O2 produces 5L of O2, the value of x is
  • a)
    0.2
  • b)
    2.0
  • c)
    1.0
  • d)
    2.5 
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
H2O2 in aqueous solution decomposes on warming as: 2H2O2(aq) → 2...
Solution:

Given: 2H2O2(aq) 2H2O(l) O2(g)

1 mole of gas occupies a volume of 25 L under the conditions of measurement.

200 ml of x M solution of H2O2 produces 5L of O2.

We need to find the value of x.

Let's calculate the number of moles of O2 produced.

Volume of O2 produced = 5 L

According to the ideal gas law, PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.

Under the given conditions, P = 1 atm, V = 5 L, T = 273 + 25 = 298 K

R = 0.0821 L atm K-1 mol-1

n = PV/RT

n = (1 atm x 5 L) / (0.0821 L atm K-1 mol-1 x 298 K)

n = 0.204 mol

From the balanced equation, we can see that 1 mole of H2O2 produces 1/2 mole of O2.

So, the number of moles of H2O2 used = 0.204 x 2 = 0.408 mol

Volume of H2O2 used = (0.408 mol x 34 mL) / (x M)

Volume of H2O2 used = 13.872 / x

Now, we can equate the two volumes:

13.872 / x = 200 / 1000

x = 2 M

Therefore, the value of x is 2.0.

Answer: Option B (2.0)
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H2O2 in aqueous solution decomposes on warming as: 2H2O2(aq) → 2...
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H2O2 in aqueous solution decomposes on warming as: 2H2O2(aq) → 2H2O(l) + O2(g) If 1 mole of gas occupies a volume of 25 L under the conditions of measurement and 200 ml of x M solution of H2O2 produces 5L of O2, the value of x isa)0.2b)2.0c)1.0d)2.5Correct answer is option 'B'. Can you explain this answer?
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