MCQ
Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are
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- A$B$ only
- B$C$ only
- ✓$B$ and $C$ only
- D$A$ and $B$ only
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$PCl _{3}+ H _{2} O \longrightarrow A + HCl$
$A + H _{2} O \longrightarrow B + HCl$
number of ionisable protons present in the product $B\,.....$
Given $\mathrm{R}=8.31\, \mathrm{~J} \,\mathrm{~K}^{-1} \,\mathrm{~mol}^{-1} ; \log 6.36 \times 10^{-3}=-2.19$ $\left.10^{-4.79}=1.62 \times 10^{-5}\right]$