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A given amount of gas occupies 10.0dm5 at 4atm and 273°C. The number of moles...


Question

A given amount of gas occupies 10.0dm5 at 4atm and 273°C. The number of moles of the gas present is [Molar volume of gas at s.t.p = 22.4dm\(^3\)]

Options

A) 0.89 mol

B) 1.90 mol

C) 3.80 mol

D) 5.70 mol


The correct answer is A.

Explanation:

PV = nRT
R = 0.08206(gas constant)
T = 273 + 273 = 546k(kelvin)
Therefore, n = PV/RT
n = 4 x 10/ 546 x 0.08206 = 0.8869, approx. 0.89mol

Explanation provided by David Gomba


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Dicussion (1)

  • PV = nRT
    R = 0.08206(gas constant)
    T = 273 + 273 = 546k(kelvin)
    Therefore, n = PV/RT
    n = 4 x 10/ 546 x 0.08206 = 0.8869, approx. 0.89mol