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Sulphur on reaction with concentrated ${HNO}_{3} .$ gives (A) which reacts with $NaOH$ gives (B). (A) and (B) are

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${NO}_{2}, {Na}_{2} S$

$H_{2} S_{2} O_{3}, {Na}_{2} S_{2} O_{3}$

$H_{2} {SO}_{4}, {Na}_{2} {SO}_{4}$

$H_{2} {SO}_{3},$ ${Na}_{2} S_{2} O_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} S + {HNO}_{3} ⟶ (A) \\ conc. \\ (A) + NaOH ⟶ (B) \\ S + 6 {HNO}_{3} ⟶ H_{2} {SO}_{4} + 6 {NO}_{2} + 2 H_{2} O \\ (A) \\ H_{2} {SO}_{4} + 2 NaOH ⟶ {Na}_{2} {SO}_{4} + 2 H_{2} O \\ (A) \end{array}$

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