Question

A 31.4 mL sample of nitrogen gas was collected over water at 23.7°C and a total pressure of 706 torr. What mass of nitrogen gas was collected

Answer 1

Answer:

$m=0.0325g$

Explanation:

Hello!

In this case, since the collection of gases over at certain conditions implies the usage of the ideal gas equation and computation of the pressure of the gas by subtracting the vapor pressure of water (21.7 torr at 23.7 °C) to the total pressure as shown below:

$p_N=706-21.7=684.3torr =0.900atm$

Next step is to compute the moles of nitrogen at the given conditions:

$n=\frac{PV}{RT}=\frac{0.900atm*0.0314L}{0.08206\frac{atm*L}{mol*K}*296.85K} \\\\n=0.00116mol$

Final step is to compute the mass by knowing that gaseous nitrogen has a molar mass of 28.02 g/mol:

$m=0.00116mol*\frac{28.02g}{1mol} \\\\m=0.0325g$

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Answer 2

The mass of nitrogen gas that was collected is 0.0336 grams.

Given the following data:

• Volume of nitrogen gas = 31.4 mL

• Temperature = 23.7°C to K = 296.7 Kelvin

• Pressure = 706 torr to atm = 0.93 atm.

Scientific data:

• Ideal gas constant, R = 0.0821 L⋅atm/mol⋅K

• Molar mass of nitrogen gas = 28.02 g/mol

To determine the mass of nitrogen gas that was collected:

First of all, we would calculate the number of moles of nitrogen gas used by using the ideal gas law equation:

$n = \frac{PV}{RT}$

Where;

• P is the pressure.

• V is the volume.

• n is the number of moles of gas.

• R is the ideal gas constant.

• T is the temperature.

Substituting the given parameters into the formula, we have;

$n = \frac{0.93 \times 0.0314}{0.0821 \times 296.7}\\\\n=\frac{0.0292}{24.36}$

n = 0.001199 moles

For the mass of nitrogen gas:

$Mass = n \times molar \;mass\\\\Mass = 0.001199 \times 28.02$

Mass = 0.0336 grams

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