Multiple choice

The pressure exerted by a mixture of $1.6\ g$ of gas $A$ and $2.2\ g$ of gas $B$ in a $10\ { dm }^{ 3 }$ flask at $25$ is: [Molar mass of gas $A = 32\ g/mol$] [Molar mass of gas $B = 44\ g/mol$]

  1. $\dfrac { RT }{ 50 }$
  2. $\dfrac { RT }{ 100 }$
  3. $\dfrac { RT }{ 10 }$
  4. $\dfrac { RT }{ 40 }$
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B Correct answer
Explanation

Gas A contributes 1.6/32 = 0.05 mol, and gas B contributes 2.2/44 = 0.05 mol, giving 0.10 mol in total. From PV = nRT with V = 10 dm^3, P = 0.10RT/10 = RT/100.

AI explanation

First, determine the number of moles for each gas by dividing the given mass by its molar mass. This gives 1.6 divided by 32, which equals 0.05 moles of gas A, and 2.2 divided by 44, which equals 0.05 moles of gas B. The total number of moles is 0.10, and applying the ideal gas law, P equals nRT divided by V, where V is 10 dm cubed. This calculates to 0.10 multiplied by RT divided by 10, resulting in RT divided by 100.

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