One mole of any Substance contains 6.022\; \; 10^{23} of Atoms/molecules of H_{2}SO_{4} present in the 100 mL of 0.02M H_{2}SO_{4} solution is

(i) 12.044 \times 10^{20} molecules

(ii) 6.022 \times 10^{23} molecules

(iii) 1 \times 10^{23}molecules

(iv) 12.044 \times 10^{23} molecules

Answers (1)

The answer is the option (i) 12.044\times 10^{20} molecules.

Explanation:

Given: Molarity = 0.02 M

Given; Volume of solution = 100ml = 0.1 L

Number of moles Of H_{2}SO_{4} = Molarity ×  Volume ( litres)

Number of molecules of H_{2}SO_{4} = 2 \times 10^{-3} \times 6.022 \times 10^{23}

Number of molecules  = 12.044\times 10^{20} Molecules

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