MCQ
The number of possible isomers for compound ${C_2}{H_3}C{l_2}Br$ is
- A$2$
- ✓$3$
- C$4$
- D$5$

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. . . . . (Given that the value of solubility product of $A B \left( K _{ sp }\right)=2 \times 10^{-10}$ and the value of ionization constant of $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )