calculate the volume of 10^21 molecules at NTP?

Avogadro constant = 6.02 × 10²³
No. of moles of the molecules = 10²¹ / (6.02 × 10²³) mol

Pressure, P = 1 atm
Volume, V = ? L
Number of moles, n = 10²¹ / (6.02 × 10²³) mol
Gas constant, R = 0.0821 L atm / (mol K)
Temperature, T =298 K (for NTP)

Gas law: PV = nRT
Then, V = nRT/P

Volume, V = [10²¹ / (6.02 × 10²³)] × 0.0821 × 298 / 1 L = 0.0406 L = 40.6 mL

pV = nRT

(100 kPa)( V ) = (10^21/(6.022 x 10^23)) * 8.314 J/(molK) * 273.15 K =>
V = 0.0000377 m^3 or 37.7 cc.

At standard temperature and pressure, the volume of one mole of a gas is 22.4 liters. The only difference between normal temperature and pressure and standard temperature is the temperature. Standard temperature is 273˚K. Normal temperature is 293˚K. An increase of temperature will increase the volume of gas. To determine the number of moles of the gas, divide the number of molecules by 6.02 * 10^23.

n = 1 * 10^21 ÷ 6.02 * 10^23
This is approximately 1.66 * 10^-3 mole.

Let’s use the following equation to determine the volume of one mole of a gas at normal temperature and pressure.

V = 22.4 * 293 ÷ 273 = 6563.2 ÷ 273

This is approximately 24.04 liter. To determine the volume of this gas, multiply the number of moles by the volume of one mole.

V = n * 22.4 * 293 ÷ 273
V = (1 * 10^21 ÷ 6.02 * 10^23) * 6563.2 ÷ 273
This is approximately 0.04 liter. I hope this is helpful for you.

At standard temperature and pressure, the volume of one mole of a gas is 22.4 liters. The only difference between normal temperature and pressure and standard temperature is the temperature. Standard temperature is 273˚K. Normal temperature is 293˚K. An increase of temperature will increase the volume of gas. To determine the number of moles of the gas, divide the number of molecules by 6.02 * 10^23.

n = 1 * 10^21 ÷ 6.02 * 10^23
This is approximately 1.66 * 10^-3 mole.

Let’s use the following equation to determine the volume of one mole of a gas at normal temperature and pressure.

V = 22.4 * 293 ÷ 273 = 6563.2 ÷ 273

This is approximately 24.04 liter. To determine the volume of this gas, multiply the number of moles by the volume of one mole.

V = n * 22.4 * 293 ÷ 273
V = (1 * 10^21 ÷ 6.02 * 10^23) * 6563.2 ÷ 273
This is approximately 0.04 liter. I hope this is helpful for you.

You have :
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Vntp = ( Vntp* ) ( n) = ( Vstp* ) ( N / NAVO ) ( Tnpt / Tstp )

Vntp = ( 22.4 L at STP/mol ) ( 1.00 x 10^21 molecules / 6.022 x 10^23 molecules/mol ) ( 293.2 K / 273.2 K )

Vntp = ( 22.4 L at NTP / mol ) ( 1.783 x 10^-3 mols )

Vntp = 39.9 x 10^-3 L at NTP = 39.9 mL at NTP <---------------------

STP = 273K and 100 kPa (although many still use 1 atm)

NTP = 293K and 1 atm

Unless you are told otherwise by your teacher/professor