# Question #f4c7f

Jun 19, 2017

$2.80 \times {10}^{23}$ molecules.

#### Explanation:

You have to be careful with the “N” ( normal temperature (20’C)) and STP ( standard 0’C, in degrees Kelvin (273.15) for the equation). One is for convenience, the other is for accuracy. The “22.4L/mol” figure is handy and useful, but based on the conditions at STP, not NTP.

Start with the Ideal Gas Law: $P \cdot V = \left(n \cdot R \cdot T\right)$
Rearranging the Ideal Gas Law equation we have $V = \frac{n \cdot R \cdot T}{P}$
where R = 0.0820575 L-atm/K-mol , P = 1.0 atm, n = 1 mol and T = 0’C = 273.15’K (STP)

$V = \frac{n \cdot R \cdot T}{P}$ ; $V = \frac{1.0 \cdot 0.0821 \cdot 273.15}{1.0}$
$V = \frac{1.0 \cdot 0.0821 \cdot 273.15}{1.0} = 22.4 L$

At NTP this would be

$V = \frac{1.0 \cdot 0.0821 \cdot 293.15}{1.0} = 24.1 L$!
That’s a 10% difference from the STP value!

Directly calculated for this problem, we have:

$n = \frac{P \cdot V}{R \cdot T}$ ; $n = \frac{1.0 \cdot 11.2}{0.0821 \cdot 293.15} = 0.465 m o l$

Applying Avogadro's Number to this we obtain: $0.465 m o l \cdot 6.022 \times {10}^{23} = 2.80 \times {10}^{23}$ molecules.