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Chapter 18: Problem 15

The average concentration of carbon monoxide in air in an Ohio city in 2006 was 3.5 ppm. Calculate the number of CO molecules in 1.0 L of this air at a pressure of 759 torr and a temperature of \(22^{\circ} \mathrm{C}\).

Short Answer

Expert verified
There are approximately \( 8.64 x 10^{19} \) carbon monoxide (CO) molecules in 1.0 L of this air in an Ohio city in 2006.

Step by step solution

01

Convert the temperature to Kelvin

To use the Ideal Gas Law, the temperature must be in Kelvin (K). To convert the temperature from Celsius to Kelvin, use the formula: \[ T(K) = T(^\circ C) + 273.15 \] Plugging in the given temperature, we get: \[ T(K) = 22^\circ C + 273.15 = 295.15 K \]
02

Convert pressure from torr to atm

The Ideal Gas Law requires pressure to be in atmospheres (atm). To convert the pressure from torr to atm, use the conversion factor (1 atm = 760 torr): \[ P(atm) = \frac{P(torr)}{760} \] Given the pressure of 759 torr, we have: \[ P(atm) = \frac{759}{760} = 0.99868 atm \]
03

Calculate moles of air in the 1.0 L volume

Now, we can plug the pressure, volume, and temperature into the Ideal Gas Law to find the moles of air in the 1.0 L volume. The Ideal Gas Law equation is: \[ PV = nRT \] Where: P is the pressure in atm (0.99868 atm), V is the volume in liters (1.0 L), n is the number of moles, R is the ideal gas constant (0.08206 L atm / K mol), and T is the temperature in Kelvin (295.15 K). Rearrange the equation to find n: \[ n = \frac{PV}{RT} \] Now, plug in the values and solve for the number of moles: \[ n = \frac{(0.99868)(1)}{(0.08206)(295.15)} = 0.04097 mol \]
04

Calculate the moles of CO in the 1.0 L volume

Since the concentration of CO is given in ppm (parts per million), we can use this ratio to find the moles of CO in the sample. The ratio is: \[ \frac{moles\ CO}{moles\ total} = \frac{3.5}{10^6} \] Rearrange the equation and plug in the values to find the moles of CO: \[ moles\ CO = \frac{3.5}{10^6} \times 0.04097 = 1.434 x 10^{-4}\ mol\ CO \]
05

Calculate the number of CO molecules

Finally, we can use Avogadro's number (6.022 x 10^23 particles/mol) to convert the moles of CO to the number of CO molecules: \[ molecules\ CO = (1.434 x 10^{-4}\ mol\ CO) (6.022 x 10^{23} particles/mol) \] \[ molecules\ CO = 8.64 x 10^{19} \] So, there are approximately \( 8.64 x 10^{19} \) carbon monoxide (CO) molecules in 1.0 L of this air in an Ohio city in 2006.

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