A) O.125M
B) 0.178M
C) 0.21 OM
D) 0.253 M
Correct Answer: A
Solution :
\[\underset{1\,\text{mol}}{\mathop{CaC{{l}_{2}}}}\,\xrightarrow{{}}C{{a}^{2+}}+\underset{2\,\times \,6.02\,\times \,{{10}^{23}}\text{ions}}{\mathop{2C{{l}^{-}}}}\,\] \[\therefore \] \[301\times {{10}^{22}}C{{l}^{-}}\]ions will be present in \[CaC{{l}_{2}}=\frac{1\times 3.01\times {{10}^{22}}}{2\times 6.02\times {{10}^{23}}}\] \[=0.025\,mol\] Molarity of solution \[=\frac{0.025\,mol}{0.200\,L}=0.125\,M\]
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