Question

2.24L of N2 at NTP contain same no of molecules are present in 1.7 gm. of ammonia prove it.​

Answer 1

Answer:

Thus proved that the same number of molecules are present in 2.24 liters of N2 at NTP and in 1.7 grams of ammonia.

Explanation:

To prove that the same number of molecules are present in 2.24 liters of N2 at NTP (Normal Temperature and Pressure) and in 1.7 grams of ammonia, we need to use the concept of Avogadro's law and the ideal gas equation.

Avogadro's law states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules.

The ideal gas equation relates the number of molecules to the mass, volume, and molar mass of a gas.

Let's calculate the number of molecules in each case:

For N2 at NTP:

Determine the molar mass of nitrogen (N2): 28 grams/mol (2 nitrogen atoms at 14 grams/mol each).

Use the ideal gas equation to calculate the number of moles:  $n = \frac{PV}{RT}$

Assuming NTP is 1 atm pressure and 273 K temperature:

$n = \frac{(1 atm * 2.24 L)}{(0.0821 atmL/molK * 273 K)}$

$n$  ≈  $0.095 mol$

Convert moles to molecules using Avogadro's number $(6.022 *10^{23} molecules/mol)$:

Number of molecules = $0.095 mol * (6.022 *10^{23} molecules/mol)$

For ammonia (NH3):

Determine the molar mass of ammonia (NH3): 17 grams/mol (1 nitrogen atom at 14 grams/mol and 3 hydrogen atoms at 1 gram/mol each).

Convert grams to moles using the molar mass:

n = mass / molar mass

n = 1.7 g / 17 g/mol

n = 0.1 mol

Convert moles to molecules using Avogadro's number:

Number of molecules = $0.1 mol * (6.022 *10^{23} molecules/mol)$

By comparing the calculated values for the number of molecules in both cases, you will find that they are equal, thus proving that the same number of molecules are present in 2.24 liters of N2 at NTP and in 1.7 grams of ammonia.

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