A) 0.2
B) 0.1
C) 0.3
D) 0.075
Correct Answer: D
Solution :
\[{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}={{M}_{3}}{{V}_{3}}\] \[0.1\times 100+2\times 0.1\times 100\,\,=\,\,x+200\] \[Conc.\,\,of\,\,N{{a}^{+}}\,=\,\,\frac{100\times 0.1}{200}+\frac{100\times 0.1\times 2}{200}\,\,=\,\,0.15\,\,M\]\[\therefore \,\,\,\,\,\,Ionic\,\,strength\,\,=\,\,\frac{1}{2}\,C{{Z}^{2}}\,=\,\frac{1}{2}\,\times \left[ 0.15\times {{1}^{2}} \right]\,\,=\,\,0.075\]You need to login to perform this action.
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