
Определить количество вещества и количество молекул в 1 кг сахара C12H22O11


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Calculation of the Amount of Substance and Number of Molecules in 1 kg of Sucrose (C12H22O11)
To determine the amount of substance and the number of molecules in 1 kg of sucrose (C12H22O11), we need to use the molar mass of sucrose and Avogadro's number.
1. Molar Mass of Sucrose (C12H22O11): The molar mass of sucrose (C12H22O11) can be calculated by summing up the atomic masses of all the atoms in the molecule. Using the atomic masses from the periodic table, we can calculate the molar mass as follows:
- Carbon (C): 12.01 g/mol (12 atoms in sucrose) - Hydrogen (H): 1.01 g/mol (22 atoms in sucrose) - Oxygen (O): 16.00 g/mol (11 atoms in sucrose)
Therefore, the molar mass of sucrose is: (12.01 g/mol * 12) + (1.01 g/mol * 22) + (16.00 g/mol * 11) = 342.3 g/mol.
2. Calculation of the Amount of Substance: The amount of substance is measured in moles. To calculate the amount of substance in 1 kg of sucrose, we can use the following formula:
Amount of Substance (in moles) = Mass (in grams) / Molar Mass (in g/mol)
Since we have the mass in kilograms, we need to convert it to grams by multiplying by 1000:
Mass of 1 kg of sucrose = 1 kg * 1000 g/kg = 1000 g
Therefore, the amount of substance in 1 kg of sucrose is: Amount of Substance = 1000 g / 342.3 g/mol = 2.92 mol
3. Calculation of the Number of Molecules: Avogadro's number (6.02 x 10^23) represents the number of particles (atoms, molecules, or ions) in one mole of a substance. To calculate the number of molecules in 1 kg of sucrose, we can use the following formula:
Number of Molecules = Amount of Substance (in moles) * Avogadro's Number
Therefore, the number of molecules in 1 kg of sucrose is: Number of Molecules = 2.92 mol * (6.02 x 10^23 molecules/mol) = 1.76 x 10^24 molecules.
In conclusion, in 1 kg of sucrose (C12H22O11), there are approximately 2.92 moles of sucrose and approximately 1.76 x 10^24 molecules of sucrose.
Please note that the values provided are approximate and may vary slightly depending on the exact atomic masses used and the level of precision required for the calculation.
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