[ANSWER] How many molecules of N2 are in a 400.0 mL container at 780 mm Hg and 135°C?

Avogadro’s number = 6.022 x 1023

A. 7.01 × 1021 molecules

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[ANSWER] How many molecules of N2 are in a 400.0 mL container at 780 mm Hg and 135°C?
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B. 7.38 × 1021 molecules

C. 2.12 × 1022 molecules

D. 2.23 × 1022 molecules

 

EXPERT ANSWER

“B. 7.38 × 1021 molecules.” To begin, the number of moles must be determined. The ideal gas equation will be utilized to do this. PV = nRT, where n = PV/RT. As per the question, the variables take on the following values: P = 780mmHG which is similar to 103991 Pa, V = 0.4L, T = 408.15K, R = 8314.463LPa/K.mol. When these values are substituted into the equation, the following results: n = (103991 x 0.4)/ (8314.463 x 408.15) = 0.012 moles. Since one mole possesses 6.02 × 1023 particles, 0.012 moles will comprise of 0.012 x 6.02 x 1023 molecules. This results in 7.38 × 1021 molecules.

To begin with, we need to find the number of moles through the help of the gas equation which is; PV = nRT therefore, n = PV/RT. We then replace with the figures from the question; P = 780mmHG, which is converted into pascal; 760mHG = 101325pa, what about 780mmHG? Therefore, (780 * 101325)/760 = 103,991 Pa. Thereafter, V = 400ml which is also same as 0.4L. T = 135C = 135 + 273.15 = 408.15K. n = ? and R = 8314.463LPa/K.mol. n= (103991 * 0.4)/(8314.463 * 408.15)= 0.012moles. we then convert moles into molecules by 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules.

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