A) 0.04
B) 0.1
C) 0.39
D) 0.62
Correct Answer: A
Solution :
[a] \[\Lambda _{eq}^{\infty }(N{{H}_{4}}OH)=\Lambda _{eq}^{\infty }(N{{H}_{4}}Cl)+\Lambda _{eq}^{\infty }(NaOH)-\Lambda _{eq}^{\infty }(NaCl)\] \[=129.8+217.8-109.3=238.3\text{ }oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}\] \[\alpha =\frac{{{\Lambda }_{eq}}}{\Lambda _{eq}^{\infty }}=\frac{9.30}{238.3}=0.04\]You need to login to perform this action.
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