1. 100 g.
2. 24 g.
3. 12.044 ×1023 oxygen atoms.
4. 6.022 * 10^21
5. 3.011 × 10^23 molecules.
6. 1000
7. 1000
8. 500g
9. 1.566 X 10^24
10.40%, 6.66% and 53.33%.
1. CaCO3 = 40+12+3×16 = 100 g
No. of moles of CaCO3 = No.ofmolecules/Avogadro constant = 6.022 × 1023/ 6.022 × 1023 = 1
mole Mass of CaCO3 = No. of moles × molar mass = 1 × 100 g = 100 g.
2. of atoms/Avogadro constant = 12.044 × 1023/6.022 × 1023 = 2
mole Mass of carbon atoms = No. of moles × atomic mass = 2 × 12 = 24 g.
3. Solution — 1 molecule of O2 = 2 oxygen atoms So, 1 mole of O2 = 2 mole oxygen atoms = 2 × 6.022 × 1023 = 12.044 ×1023 oxygen atoms.
4.The atomic mass of Copper ( Cu ) is 63.5 u . There fore , the molar mass of Copper is 63.5 g . So , 63.5 g of Cu will contain 6.022 * 10^23 . So , 0.635 g of Cu will contain 6.022 * 10^23 / 100 = 6.022 * 10^21
5.Hence, 11.2 liters of SO2 gas at NTP contains 3.011 × 10^23 molecules.
6.Atomic mass of the element =Massofoneatom×6.02×10²³
=6.644×10
−23
×6.02×10
23
=40g
No. of gram atoms = No.ofgram
MMassoftheelementingram
= 40
40000
=1000
7. seym lang sa number
9. there are 1.566 X 10^24 atoms of oxygen in 0.2 moles of Na2CO3. 10H2O
10. elements present in are C, Oxygen and H. The atomic mass of Carbon, Hydrogen and Oxygen are 12,1 and 16. The percentage of composition of C, H and O in glucose is 40%, 6.66% and 53.33%.
