Equivalent mass of $Na X_{2} CO X_{3}$ (molar mass $106 g mo l^{- 1}$ ) when it reacts completely with excess $HCl$ to give $CO X_{2}$ is:

Answer: B

Na X_{2} CO X_{3} + 2 HCl 2 NaCl + H X_{2} O + CO X_{2}

: one mole consumes two moles of

H X^{+}

, so

n = 2

. Equivalent mass

= 106/2 = 53

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