A) 10.56 gm
B) 15 gm
C) 12.74 gm
D) 16.25 gm
Correct Answer: A
Solution :
\[pOH=pKb+\log \frac{[Salt]}{[Base]}\] \[14-9.35=-\log (1.78\times {{10}^{-5}})+\log \frac{[Salt]}{100}\] \[[Salt]=79.9\]Þ \[\frac{w}{132}\times 1000=79.9\]Þ \[w=10.56\]You need to login to perform this action.
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