50 mL of 1 M oxalic acid is shaken with 0.5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The initial moles of oxalic acid $= M_1 	imes V = 1 \ M 	imes 0.05 \ L = 0.05 \ mol$.The final moles of oxalic acid $= M_2 	imes V = 0.5 \ M

Vedclass · Accepted Answer

$6.30$

Vedclass · Answer

(C) The initial moles of oxalic acid $= M_1 	imes V = 1 \ M 	imes 0.05 \ L = 0.05 \ mol$.The final moles of oxalic acid $= M_2 	imes V = 0.5 \ M 	imes 0.05 \ L = 0.025 \ mol$.The moles of oxalic acid adsorbed $= 0.05 - 0.025 = 0.025 \ mol$.The molar mass of oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.The mass of oxalic acid adsorbed by $0.5 \ g$ of charcoal $= 0.025 \ mol 	imes 126 \ g/mol = 3.15 \ g$.Therefore,the amount of oxalic acid adsorbed per $g$ of charcoal $= \frac{3.15 \ g}{0.5 \ g} = 6.30 \ g/g$.

List-$I$	List-$II$
$A$. Physisorption	$I$. Single layer adsorption
$B$. Chemisorption	$II$. $20-40 \, kJ \, mol^{-1}$
$C$. $N_{2(g)} + 3H_{2(g)} \xrightarrow{Fe(s)} 2NH_{3(g)}$	$III$. Chromatography
$D$. Analytical Application of Adsorption	$IV$. Heterogeneous catalysis

$50 \ mL$ of $1 \ M$ oxalic acid is shaken with $0.5 \ g$ of wood charcoal. The final concentration of the solution after adsorption is $0.5 \ M$. The amount of oxalic acid adsorbed per $g$ of charcoal is ....... $g$.

Similar Questions

Identify the $FALSE$ statement regarding adsorption from the following.

If $x$ is the amount of adsorbate and $m$ is the amount of adsorbent,which of the following relations is not related to the adsorption process?

In the figure,a straight line is given for the Freundlich Adsorption isotherm equation $(y = mx + c)$,where the equation is $y = 0.3x + 0.7033$. The values of $\frac{1}{n}$ and $\log K$ are respectively:

Consider the graphs $(I)$,$(II)$ and $(III)$. The correct statement related to the graphs is/are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module