A) \[BaC{{l}_{2}}.3{{H}_{2}}O\]
B) \[BaC{{l}_{2}}.2{{H}_{2}}O\]
C) \[BaC{{l}_{2}}.{{H}_{2}}O\]
D) \[BaC{{l}_{2}}.4{{H}_{2}}O\]
Correct Answer: B
Solution :
Weight of Hydrated \[BaC{{l}_{2}}=61g\] Weight of Anhydrous \[BaC{{l}_{2}}=52g\] Loss in mass \[=9g\] Assuming \[BaC{{l}_{2}}.x{{H}_{2}}O\] as hydrate Mass of \[{{H}_{2}}O=9g\] Moles of \[{{H}_{2}}O=\frac{9}{18}=0.5\] Grass molecular let of \[BaC{{l}_{2}}=208\]% of \[{{H}_{2}}O\] in this hydrated \[BaC{{l}_{2}}=\frac{9}{61}\times 100=14.75%\] \[=\frac{18x}{208+18x}\times 100\]on solving \[x=2\] This percentage is present in \[BaC{{l}_{2}}.2{{H}_{2}}O\]
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