Consider the ionisation of H₂SO₄ as follow:
$H_{2} {SO}_{4} + 2 H_{2} O \underset{}{\to} 2 H_{3} O^{\oplus} + {SO}_{4}^{2 -}$
The total number of ions furnished by 100 mL of 0.1 M H₂SO₄ will be

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1.2 $\times$ 10²³

0.12 $\times$ 10²³

1.8 $\times$ 10²³

0.18 $\times$ 10²³

answer is C.

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Detailed Solution

$1 mol H_{2} {SO}_{4} \equiv 2 mol H_{3} O^{\oplus} \equiv 1 mol {SO}_{4}^{2 -}$

$(\frac{100}{1000} \times 0.1 = 0.01) \equiv 2 \times 0.01 + 0.01 = 0.03 mol$

Total number of g ions = 0.03 x 6.02 x 10²³

=0.18069 x 10²³

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