A) 4.25 M
B) 0.425 M
C) 42.5 M
D) 0.74 M
Correct Answer: B
Solution :
Given, \[{{\operatorname{M}}_{1}} = 0.26 M, {{M}_{2}} = 0.48 M\] \[{{\operatorname{V}}_{1}}=50mL,\,{{V}_{2}}=\,\,150\,mL\] Let \[{{\operatorname{M}}_{3}}\,\,=\,\,molarity\,\,of\,\,resultant\,\,solution\] \[{{V}_{3}}=150+50\] \[= \,\,200 mL\] Applying mixing formula \[{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}={{M}_{3}}{{V}_{3}}\] \[{{M}_{3}}=\frac{{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}}{{{V}_{3}}}\] \[=\,\,\,\frac{0.26\times 50+0.48\times 150}{200}\] \[= \,0.425 M\]You need to login to perform this action.
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