Calculating the Mass and Number of Molecules in 120 cc of Nitrogen Gas at NTP

In this article, we will delve into the process of calculating the mass and number of molecules present in 120 cc of nitrogen gas at Normal Temperature and Pressure (NTP). Understanding these fundamental concepts is crucial in various scientific and industrial applications. We will walk through each step, from understanding the conditions of NTP to determining the number of molecules using Avogadro's number.

Understanding NTP Conditions

The standard conditions for Normal Temperature and Pressure (NTP) are defined as:

Temperature: 0°C (273.15 K) Pressure: 1 atm (101.325 kPa)

At these conditions, 1 mole of an ideal gas occupies approximately 22.4 liters or 22,400 cc.

Molar Volume of a Gas at NTP

Step 2: Molar Volume of a Gas at NTP

At NTP, 1 mole of an ideal gas occupies approximately 22.4 liters or 22,400 cc. This volume is known as the molar volume of the gas at NTP.

Calculating Moles of Nitrogen

Now, let's calculate the number of moles of nitrogen in 120 cc:

(text{Moles of } N_2 frac{text{Volume of } N_2}{text{Molar Volume at NTP}} frac{120 , text{cc}}{22400 , text{cc/mol}} approx 0.00536 , text{moles})

Calculating the Mass of Nitrogen

The molar mass of nitrogen ((N_2)) is approximately 28 g/mol. We can use the molar mass to determine the mass of 120 cc of nitrogen:

(text{Mass of } N_2 text{Moles of } N_2 times text{Molar Mass of } N_2 0.00536 , text{moles} times 28 , text{g/mol} approx 0.150 , text{g})

Calculating the Number of Molecules

To find the number of molecules, we use Avogadro's number, which is approximately (6.022 times 10^{23}) molecules/mol:

(text{Number of molecules} text{Moles of } N_2 times text{Avogadros Number} approx 0.00536 , text{moles} times 6.022 times 10^{23} , text{molecules/mol} approx 3.22 times 10^{21} , text{molecules})

Summary

Mass of 120 cc of nitrogen at NTP: Approximately 0.150 g Number of molecules present: Approximately (3.22 times 10^{21}) molecules

Additional Calculations

Alternatively, you can perform the calculation in a simpler manner:

No. of N2 molecules present in 22,400 cc: 6.023 × 1023 No. of molecules present in 120 cc: (frac{120 , text{cc} times 6.023 times 10^{23}}{22400 , text{cc}} 3.21 times 10^{21})

Molar Mass of Nitrogen

The molar mass of (N_2) is 28 g/mol, which means 28 grams of (N_2) occupy 22.4 L (22,400 cc) at NTP.

Volume of the Gas and Mass Calculation

Volume of the gas 120 mL 0.12 L

Volume of 1 mole (28 g) of nitrogen at STP 22.4 L

Mass of 0.12 L of nitrogen at STP (text{Mass} 28 , text{g} times frac{0.12 , text{L}}{22.4 , text{L}} 0.15 , text{g})

Number of Molecules in 0.15 g of Gas

The number of molecules in 1 mole (28 g) of gas is (6.022 times 10^{23}) molecules.

(text{Number of molecules in 0.15 g of gas})

(text{Number of molecules} frac{6.022 times 10^{23} times 0.15 , text{g}}{28 , text{g}} 3.22 times 10^{21} , text{molecules})

Conclusion

In this article, we have thoroughly analyzed the process of calculating the mass and number of molecules in 120 cc of nitrogen gas at NTP. The importance of understanding these concepts is highlighted, and the step-by-step calculations are provided to ensure clarity and accuracy in scientific and industrial applications.