Question

102. 6.02 ×1020 molecules of glucose are
present in 100 mL of its solution. The
concentration of this glucose solution is
(a) 0.1 M (b) 0.2 M
(c) 0.02 M (d) 0.01 M

Answer 1

Answer:0.01

Explanation:No. of moles of glucose= 6.02x10^20÷6.02×10^23=0.001

Volume=100 mL= 0.1 L

Molarity= no. of mol/ vol

= 0.001/0.1 = 0.01

Answer 2

(d) 0.01M is the best answer.

Explanation:

• It is described because of the wide variety of moles of the solute in 1liter of the solution.
• Number moles of the given problem is = 6.02 × 10²⁰.
• Number of moles =  6.02 × 10²⁰/  6.02 × 10²³ = 10⁻³mol.
• Volume of the solution =100ml =0.1L.

Therefore, we know, Molarity = number of moles/Volume = 10⁻³/0.1 = 0.01M.

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