A) 0.8 M
B) 1.6 M
C) 1.2M
D) 10.0 M
Correct Answer: B
Solution :
1 mol of \[HCl\] give 1 mol \[C{{l}^{-}}\] \[\therefore \] 0.4 mol of \[HCl\]will give \[0.4\text{ }mol\,C{{l}^{-}}\] 1 mol \[CaC{{l}_{2}}\] give \[2\text{ }mol\text{ }C{{l}^{-}}\] \[\therefore \] \[0.2\text{ }mol\text{ }CaC{{l}_{2}}\] give \[2\times 0.2=0.4\text{ }mol\text{ }C{{l}^{-}}\] Total moles of \[C{{l}^{-}}=0.4+0.4=0.8\text{ }mol\] Molarity \[=\frac{moles}{V(L)}=\frac{0.8}{0.5}=1.6M\]
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