Moles and Molar Mass Practice

Sample questions

5 of 7 — sign in to practice the rest with adaptive difficulty and mastery tracking.

1. Sample 1difficulty 1/5

   Avogadro's number, the number of particles in one mole, is

   • A

     $6.02 \times 10^{22}$

   • B

     $6.02 \times 10^{23}$

     check_circle
   • C

     $3.01 \times 10^{23}$

   • D

     $6.02 \times 10^{24}$

   Why

   $N_\text{A} = 6.022 \times 10^{23}\,\text{mol}^{-1}$.

   View full solution & statsarrow_forward

2. Sample 2difficulty 2/5

   How many moles are in 88 g of CO₂?

   • A

     2 mol

     check_circle
   • B

     4 mol

   • C

     0.5 mol

   • D

     1 mol

   Why

   $n = m/M = 88/44 = 2$ mol.

   View full solution & statsarrow_forward

3. Sample 3difficulty 2/5

   How many atoms are in 0.50 mol of helium?

   • A

     $0.50 \times 10^{23}$

   • B

     $1.20 \times 10^{24}$

   • C

     $3.01 \times 10^{23}$

     check_circle
   • D

     $6.02 \times 10^{23}$

   Why

   $0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ atoms.

   View full solution & statsarrow_forward

4. Sample 4difficulty 2/5

   The molar mass of CO₂ is approximately

   • A

     44 g/mol

     check_circle
   • B

     28 g/mol

   • C

     32 g/mol

   • D

     12 g/mol

   Why

   C: 12 + 2 × O: 16 = 12 + 32 = 44 g/mol.

   View full solution & statsarrow_forward

5. Sample 5difficulty 2/5

   How many oxygen atoms are in 0.25 mol of CO₂?

   • A

     $3.0 \times 10^{22}$

   • B

     $3.0 \times 10^{23}$

     check_circle
   • C

     $1.5 \times 10^{23}$

   • D

     $6.0 \times 10^{23}$

   Why

   0.25 mol CO₂ × 2 O / molecule = 0.50 mol O atoms × $6.02 \times 10^{23}$ = $3.01 \times 10^{23}$.

   View full solution & statsarrow_forward