$50 m L$ of $1 M$ oxalic acid (molar mass $= 126$ ) is shaken with $0.5 g$ of wood charcoal. The final concentration of the solution after adsorption is $0.5 M$ . What is the amount of oxalic acid adsorbed per $g r a m$ of carbon?

3.15

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1.575

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6.30

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12.60

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Solution

The correct option is C $6.30$
Initial Molarity of Oxalic Acid = 1M
Final Molarity of Oxalic Acid = 0.5M

We know, Molarity = $\frac{M o l e s}{V o l u m e}$

Moles = $M o l a r i t y \times V o l u m e$

Hence, Initial number of molesc=> $m_{i}$ = $1 M \times 50 m L$

=> $m_{i}$ = $50 m - m o l e s$ = $50 \times 10^{- 3} M o l e s$

Similarly,
Final number of moles ( $m_{f}$ ) = $0.5 M \times 50 m L$
=> $m_{f}$ = $25 m - m o l e s$ = $25 \times 10^{- 3} M o l e s$

Moles of oxalic acid adsorbed $= (50 - 25)$ m-moles $= 25$ m-moles

Moles of oxalic acid adsorbed per gram Carbon = $\frac{25 \times 10^{- 3} M o l e s}{0.5 g}$ $= 0.05$ moles

Mass of oxalic acid adsorbed = $0.05 \times 126$ $= 6.30 g$

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$50 m L$ of $1 M$ oxalic acid (molar mass $= 126$ ) is shaken with $0.5 g$ of wood charcoal. The final concentration of the solution after adsorption is $0.5 M$ . What is the amount of oxalic acid adsorbed per $g r a m$ of carbon?