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You want to determine the benzene concentration in a lab by taking a grab sample using a Mylar bag. You collect 5 liters of air from the laboratory, where the temperature was measured to be 35 C (a really hot lab!) and atmospheric pressure was 740 mmHg. Upon analyzing the sample with GC at 22 C and an atmospheric pressure of 765 mmHg the concentration was found to be 10,200 mg/m3 What was the true concentration in the laboratory measured by this method

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Final answer:

The true concentration of benzene in the lab measured by this method is 1093.54 mg/L.

Step-by-step explanation:

To determine the true concentration of benzene in the laboratory, we can use the ideal gas law and the concept of partial pressure. First, we need to convert the temperature to Kelvin and the pressure to pascals.

The temperature in Kelvin is 35 + 273 = 308 K,

and the pressure in pascals is 740 mmHg * 133.322 = 98621.48 Pa.

Next, we can use the equation:

P1/T1 = P2/T2

Where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature. We can rearrange this equation to solve for P2:

P2 = P1 * T2 / T1

Plugging in the values, we get:

P2 = 98621.48 * (295.15 / 308)

= 94289.19 Pa

This is the final pressure of the benzene in the Mylar bag. Now, we can use the ideal gas law to find the number of moles of benzene:

n = (P * V) / (R * T)

Where n is the number of moles, P is the pressure, V is the volume, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin. Plugging in the values, we get:

n = (94289.19 * 0.005) / (0.0821 * 295.15)

= 0.07 mol

Finally, we can calculate the concentration of benzene in the lab:

concentration = mass / volume

Given that the concentration is 10,200 mg/m3 and we collected 5 liters of air, we can calculate the mass of benzene:

mass = concentration * volume

= 10200 * 5

= 51000 mg

Using the molar mass of benzene (78.11 g/mol):

mass = 0.07 mol * 78.11 g/mol

= 5.4677 g

Therefore, the true concentration of benzene in the laboratory measured by this method is:

concentration = mass / volume

= 5.4677 g / 5 L

= 1.09354 g/L

= 1093.54 mg/L

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