find out which of the following has maximum number of particles:
a) 1g of Na
b) 1g of li
c) 1g of Cl2
(SOLVE COMPLETELY)
Answers
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ANSWER
1. Molecular weight of Au = 196.96 g/mol
Thus, 1g /196.96 g/mol = 0.005 mole of Au
1 mole has 6.022 � 1023 atoms (Avogadro's Number)
Thus, 0.005 mole Au is (0.005 mol x 6.022 � 1023 atoms/mol) = 3.057 x 1021 atoms
2. MW of Na = 23 g/mole
No of moles of Na = 1/23
So, atoms of Na in moles = 0.04 x 6.022 X 1023 = 2.4 x 1022 atoms
3. MW of Li = 6.94
Moles = 1/ 6.94 moles = 0.144
No of atoms = 0.144 x 6.022 X 1023 = 8.6 x 1022 atoms
4. MW of Cl2 = 71 g/moles
Moles = 1 / 71 = 0.0140
No of molecules of Cl2 = 0.0140 x 6.022 X 1023 = 8.4 x 1021
In 1 molecule of chlorine there are 2 atoms, So,
No of atoms = 2 x 8.4 x 1021 = 16.8 x 1021 = 1.68 x 1022
So, largest no of atoms are in 1 g Li
OR,
Answer
(i) Gram atomic mass of Au= 197 g
Or
197g of Au contains = 6.022 x 1023
Therefore 1gm of Au contains = 6.022 x 1023/197*1 = 3.06 x 1021 atoms
(ii) Gram atomic mass of Na = 23 g
Or
23 g of Na contains atoms = 6.022x 1023
Or
1gm of Na contains atoms = 6.022x1023 /23 *1 = 26.2 x1021 atoms
(iii) Gram atomic mass of Li = 7
Or
7g of Li contains atoms = 6.022 x 1023
Or
1g of Li contains atoms = 6.022 x 1023/7 *1= 86.0 x 1021 atoms
(iv) Gram atomic mass of Cl = 71 Or 71g of Cl contains atoms = 6.022x1023
Or
1 g of Cl contains atoms = 6.022x1023 /71 * 1= 8.48 x 1021 atoms
Hence, 1 g of Li (s) will have the largest number of atom
EXPLANATION
How to Find the Number of Representative Particles in Each Substance
1.Measure Mass. Measure the mass of the substance in grams. ...
2.Calculate Molar Mass. Calculate the molar mass of the substance. ...
3.Divide Mass by Molar Mass. Divide the mass measured in Step 1 by the molar mass determined in Step 2. ...
4.Multiply by Avogadro's Number.
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