A) 0.925 g
B) 0.0125 g
C) 0.25 g
D) 1 g
Correct Answer: A
Solution :
Key Idea: Use the following formulas to find the answer (i) \[pH=-\log [{{H}^{+}}]\] (ii) \[[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}\] (iii) \[N=\frac{v}{equivalent\text{ }wt\times volume}\] Given, V = 250mL = 250/1000 L \[pH=13\] \[Ca{{(OH)}_{2}}\]molecular weight \[=40+17\times 2=74\] Acidity of \[Ca{{(OH)}_{2}}=2\] \[\therefore \] Equivalent weight of \[Ca{{(OH)}_{2}}=\frac{molecular\text{ }weight}{acidity}\] \[=\frac{74}{2}\] = 37 \[pH=-\log [{{H}^{+}}]\] or \[13=-\log \,[{{H}^{+}}]\] \[\therefore \] \[[{{H}^{+}}]={{10}^{-13}}\] \[[O{{H}^{-}}]=\frac{{{10}^{-14}}}{[{{H}^{+}}]}\] \[=\frac{{{10}^{-14}}}{{{10}^{-13}}}\] \[={{10}^{-13}}\] \[={{10}^{-1}}\] = 0.1 \[\therefore \] \[N=0.1\] \[N=\frac{w}{Eq.\,wt.\,\times V}\] or \[w=N\times Eq.\,wt.\,\,\times V\] \[=0.1\times 37\times \frac{250}{1000}=0.925\,g\]
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