Explanation
Given:
The initial mass of charcoal$$ = 3g$$
Vol. of acetic acid =$$50mL$$ Normality$$ = 0.06N$$
Strength of filtrate after reaction$$ = 0.042N$$
We know,
Molarity of acetic acid soln$$ = \cfrac{{Wt/M}}{{Vol\ of\ so{l^n}}}$$
$$ \Rightarrow Wt = 0.06 \times 60 \times 0.05 = 0.180g$$
Also, normality after reaction, $$0.042 = \cfrac{W}{{60 \times 0.05}}$$
$$W = 0.126g$$
Change in mass$$ = 0.180 - 0.126 = 0.054g = 54mg$$
Amount of charcoal available$$ = 3g$$
Amount of acetic acid absorbed $$ = \cfrac{{54mg}}{{3.0g}} \times 1.0g = 18g$$
Correct option is A.
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