
The mass of oxalic acid crystals $\left( {{H_2}{C_2}{O_4}.2{H_2}0} \right)$ required to prepare 50 ml of a 0.2N solution is (Assume both ${H^ + }$ are replaced in the reaction)
A. 4.5 g
B. 6.3 g
C. 0.63 g
D. 0.45 g
Answer
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Hint- In order to deal with this question we will use the formula for gram equivalent which is given as the ratio of molecular weight and basicity. First we will find the molecular weight. Then we will proceed further by using simple mathematical operations.
Complete answer:
Formula used- ${\text{Gram equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{basicity}}}}$
Oxalic acid exists as the dihydrate $\left( {{H_2}{C_2}{O_4}.2{H_2}0} \right)$
Molecular weight of oxalic acid dihydrate
$
= \left( {2 \times {\text{Atomic mass of }}C} \right) + \left( {6 \times {\text{Atomic mass of }}H} \right) + \left( {6 \times {\text{Atomic mass of }}O} \right) \\
= \left( {2 \times 12g/mol} \right) + \left( {6 \times 1g/mol} \right) + \left( {6 \times 16g/mol} \right) \\
= 24g/mol + 6g/mol + 96g/mol \\
= 126g/mol \\
$
Gram equivalent weight of oxalic acid dihydrate is:
As we know that
$
\because {\text{Gram equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{basicity}}}} \\
\Rightarrow {\text{Gram equivalent weight}} = \dfrac{{{\text{126g/mol}}}}{{\text{2}}} \\
\Rightarrow {\text{Gram equivalent weight}} = {\text{63g/mol}} \\
$
So, 1 Normal solution of the oxalic acid must be having 63 g content in 1 L.
0.2 Normal solution of the oxalic acid must be having $\dfrac{{63g}}{1} \times 0.2$ = 12.6 g content in 1 L.
0.2 Normal solution of oxalic acid should have $\dfrac{{12.6g}}{{1000}} \times 50$= 0.63g in 50 mL.
Hence, 0.63 g of oxalic acid crystals is required to prepare 50 ml of a 0.2N solution.
So, the correct answer is option C.
Note- In order to solve such types of problems students must remember the formula for molecular weight and must know the method to find the atomic weight of any compound. Oxalic acid serves as a reducing agent for the elimination of tarnish and rust or ink stains from metal oxides. This is used as a laundry cleaner, wood-bleaching agent, and calcium remover for industrial use. Oxalic acid can remove some paints and varnishes. It is also used as a mordant in dyeing textiles and in the manufacture of ink.
Complete answer:
Formula used- ${\text{Gram equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{basicity}}}}$
Oxalic acid exists as the dihydrate $\left( {{H_2}{C_2}{O_4}.2{H_2}0} \right)$
Molecular weight of oxalic acid dihydrate
$
= \left( {2 \times {\text{Atomic mass of }}C} \right) + \left( {6 \times {\text{Atomic mass of }}H} \right) + \left( {6 \times {\text{Atomic mass of }}O} \right) \\
= \left( {2 \times 12g/mol} \right) + \left( {6 \times 1g/mol} \right) + \left( {6 \times 16g/mol} \right) \\
= 24g/mol + 6g/mol + 96g/mol \\
= 126g/mol \\
$
Gram equivalent weight of oxalic acid dihydrate is:
As we know that
$
\because {\text{Gram equivalent weight}} = \dfrac{{{\text{Molecular weight}}}}{{{\text{basicity}}}} \\
\Rightarrow {\text{Gram equivalent weight}} = \dfrac{{{\text{126g/mol}}}}{{\text{2}}} \\
\Rightarrow {\text{Gram equivalent weight}} = {\text{63g/mol}} \\
$
So, 1 Normal solution of the oxalic acid must be having 63 g content in 1 L.
0.2 Normal solution of the oxalic acid must be having $\dfrac{{63g}}{1} \times 0.2$ = 12.6 g content in 1 L.
0.2 Normal solution of oxalic acid should have $\dfrac{{12.6g}}{{1000}} \times 50$= 0.63g in 50 mL.
Hence, 0.63 g of oxalic acid crystals is required to prepare 50 ml of a 0.2N solution.
So, the correct answer is option C.
Note- In order to solve such types of problems students must remember the formula for molecular weight and must know the method to find the atomic weight of any compound. Oxalic acid serves as a reducing agent for the elimination of tarnish and rust or ink stains from metal oxides. This is used as a laundry cleaner, wood-bleaching agent, and calcium remover for industrial use. Oxalic acid can remove some paints and varnishes. It is also used as a mordant in dyeing textiles and in the manufacture of ink.
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