Question

1) A mole is “counting number”. The following are also counting numbers. What number do each of the following terms represent?

dozen______________
pair________________
mole_______________

2. List the three types of “representative particles” that are referred to in your notes? Explain what each are.




How many atoms are in 0.250 mol silver?


How many formula units are in 8.56 x 10-3 mol NaCl


How many molecules are in 35.4 mol CO2


How many molecules are in 0.425 mol N2


5. How many moles are in 3.25 x 1020 atoms Pb



6. How many moles are in 4.96 x 1024 molecules glucose


7. How many moles are in 1.56 x 1023 formula units NaOH


8. How many moles are in 1.25 x 1025 Cu atoms


9. How many moles are in 1.51 x 1015 atoms Si


10. How many molecules are in 4.25 x 10-2 mol NO2


11. How many moles are in 8.95 x 1025 molecules CCl4


12. How many atoms are in 5.90 mol Ca

Answer 1

1)
- dozen = 12
- pair = 2
- mole = 6.022 × 10^23 (representative particles)

2) Three types of representative particles
- Atoms — single atoms of an element are the representative particle for elemental/atomic substances (e.g., He, Fe).
- Molecules — two or more atoms covalently bonded; the representative particle for molecular compounds (e.g., O2, H2O, CO2).
- Formula units — the simplest whole-number ratio of ions in an ionic compound; the representative particle for ionic substances (e.g., NaCl, CaF2).

Use Avogadro’s number NA = 6.022 × 10^23 for conversions.

Calculations (N = n·NA, n = N/NA):

3) Atoms in 0.250 mol Ag
N = 0.250 · 6.022×10^23 = 1.51×10^23 atoms

4) Formula units in 8.56×10^-3 mol NaCl
N = 8.56×10^-3 · 6.022×10^23 = 5.15×10^21 formula units

5) Molecules in 35.4 mol CO2
N = 35.4 · 6.022×10^23 = 2.13×10^25 molecules

6) Molecules in 0.425 mol N2
N = 0.425 · 6.022×10^23 = 2.56×10^23 molecules

7) Moles in 3.25×10^20 atoms Pb
n = 3.25×10^20 / 6.022×10^23 = 5.40×10^-4 mol

8) Moles in 4.96×10^24 molecules glucose
n = 4.96×10^24 / 6.022×10^23 = 8.24 mol

9) Moles in 1.56×10^23 formula units NaOH
n = 1.56×10^23 / 6.022×10^23 = 0.259 mol

10) Moles in 1.25×10^25 Cu atoms
n = 1.25×10^25 / 6.022×10^23 = 20.8 mol

11) Moles in 1.51×10^15 atoms Si
n = 1.51×10^15 / 6.022×10^23 = 2.51×10^-9 mol

12) Molecules in 4.25×10^-2 mol NO2
N = 4.25×10^-2 · 6.022×10^23 = 2.56×10^22 molecules

13) Moles in 8.95×10^25 molecules CCl4
n = 8.95×10^25 / 6.022×10^23 = 1.49×10^2 mol (149 mol)

14) Atoms in 5.90 mol Ca
N = 5.90 · 6.022×10^23 = 3.55×10^24 atoms

If you want, I can show the intermediate arithmetic for any problem or redo any answers with different significant‑figure rules.