How many molecules of N₂ are in a 200.0 mL container at 780 mm Hg and 135°C? A) 3.50 × 10^{21 molecules} B) 3.69 × 10^{21 molecules} C) 1.06 × 10^{22 molecules} D) 1.12 × 10^{22 molecules} E) 3.30 × 10^{21 molecules}

Question

Quizplus · Accepted Answer

To solve this problem, we can use the ideal gas law equation: PV = nRT
Where P = pressure, V = volume, n = moles of gas, R = ideal gas constant, and T = temperature
First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15: 135°C + 273.15 = 408.15 K
Next, we can rearrange the ideal gas law equation to solve for n: n = PV/RT
We are given P = 780 mm Hg and V = 200.0 mL, but R depends on the units we use for pressure and volume. Since we are given pressure in mm Hg, we will use R = 62.36 L·mmHg/mol·K. (If we were given pressure in atm, we would use R = 0.0821 L·atm/mol·K.)
Plugging in the values: n = (780 mm Hg)(0.2000 L)/(62.36 L·mmHg/mol·K)(408.15 K) = 3.69 × 10^21 molecules
Therefore, the answer is B.

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