This value is 21,85 g.
Molecules of ammonia? Will assume so. 4.2 X 1025 molecules NH3 (1 mole NH3/6.022 X 1023)(17.034 grams/1 mole NH3) = 1188 grams of ammonia ===================( could call it 1200 grams NH3 for significant figure correctness )
7,02 g ammonia
1100 Grams. or 2.42 pounds.
1.025 kg is equal to 1025 grams.
Chlorine gas (at standard temperature and pressure) consists of diatomic molecules. Therefore, in the specified number of molecules of chlorine gas there are 1.364 X 1025 atoms. The gram atomic mass of chlorine, which by definition consists of Avogadro's Number of atoms, is 35.453. Therefore, the mass of the specified number of molecules of chlorine gas is 35.453 X [(1.364 X 1025)/(6.022 X 1023)] or 803 grams, to the justified number of significant digits.
The gram molecular mass of C2H4 is 28.05. Therefore, 1.26 X 103 grams constitutes (1.26/28.05) X 103 or about 44.9 moles. Each mole contains Avogadro's number of molecules; therefore, 1.26 X 103 grams contains 2.71 X 1025 molecules, to the justified number of significant digits.
There are approx 2.05*1025 molecules.
3.05 x 1025
123
A cup of coffee (200 cm3) contains about 1.5 * 1025 molecules.
A cubic meter of gas at standard temperature and pressure will have approximately 2.6 x 1025 molecules. This is based on the Avogadro's Number of molecules, (approximately 6.022 x 1023) taking up a volume of around 23 liters. Alternatively, 32 grams of oxygen has Avogadro's number of molecules.
By definition, Avogadro's Number is the number of molecules or formula units in the number of grams corresponding to the gram molecular weight or gram formula unit, and by experiment, Avogadro's Number is about 6.022 X 1023. Therefore, 7.75 X 1025 formula units contains (7.75 X 1025/6.022 X 1023) or 1.29 X 102 moles or formula units, to the justified number of significant digits.