A) 4.25 g
B) 42.5g
C) 17g
D) 1.7g
Correct Answer: A
Solution :
We know that \[m=\frac{EVN}{1000}\] Where, m = wt. E = Eq. wt. V = Volume in mL N = Normality For \[\text{AgN}{{\text{O}}_{\text{3}}}\]molarity equals to normality then, For \[\text{AgN}{{\text{O}}_{3}}\]mol. wt. = Eq. wt. \[E=170\] \[V=100\,\,mL\] \[N=0.25\] \[m=\frac{170\times 100\times 0.25}{1000}=4.25\,g\]You need to login to perform this action.
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