A) \[0.225g\]
B) \[0.0125\text{ }g\]
C) \[0.25\text{ }g\]
D) \[1\text{ }g\]
Correct Answer: A
Solution :
We know that, \[[{{H}^{+}}]={{10}^{-pH}}={{10}^{-13}}\] \[\therefore \] \[[O{{H}^{-}}]=\frac{{{10}^{-14}}}{{{10}^{-13}}}={{10}^{-1}}\] hence, normality of solution = 0.1 We know that, \[\omega =\frac{NEV}{1000}\] \[\therefore \] \[\omega =\frac{0.1\times 74\times 250}{2\times 1000}\] \[=0.925g\]
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