(B) Crystalline oxalic acid is $H_2C_2O_4 \cdot 2H_2O$. Molecular weight of $H_2C_2O_4 \cdot 2H_2O = (2 \times 1) + (2 \times 12) + (4 \times 16) + 2

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(B) Crystalline oxalic acid is $H_2C_2O_4 \cdot 2H_2O$. Molecular weight of $H_2C_2O_4 \cdot 2H_2O = (2 \times 1) + (2 \times 12) + (4 \times 16) + 2 \times (2 \times 1 + 16) = 2 + 24 + 64 + 36 = 126$. Basicity of oxalic acid is $2$. Equivalent weight $= \frac{\text{Molecular weight}}{\text{Basicity}} = \frac{126}{2} = 63$.

Class 11 Chemistry Some Basic Concepts of Chemistry Atomic, Molecular and Equivalent masses

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Equivalent weight of crystalline oxalic acid is

A
$30$
B
$63$
C
$53$
D
$45$

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Some Basic Concepts of Chemistry

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Atomic, Molecular and Equivalent masses

Molecular weight of an organic acid is given by

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Calculate the average atomic mass of chlorine using the following data:

Isotope Natural abundance (%) Molar mass $(u)$

$^{35}Cl$ $75.77$ $34.9689$

$^{37}Cl$ $24.23$ $36.9659$
(in $u$)

Isotope	Natural abundance (%)	Molar mass $(u)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

Medium

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The equivalent weight of phosphoric acid $(H_3PO_4)$ in the reaction $NaOH + H_3PO_4 \to NaH_2PO_4 + H_2O$ is

MediumAIIMS 1999

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In terms of energy,$1 \ amu$ is equal to:

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Upon fully dissolving $2.0 \ g$ of a metal in sulphuric acid,$6.8 \ g$ of the metal sulphate is formed. The equivalent weight of the metal is $...... \ g$.

MediumKVPY 2012

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Equivalent weight of crystalline oxalic acid is

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Molecular weight of an organic acid is given by

Calculate the average atomic mass of chlorine using the following data: Isotope Natural abundance (%) Molar mass $(u)$ $^{35}Cl$ $75.77$ $34.9689$ $^{37}Cl$ $24.23$ $36.9659$ (in $u$)

The equivalent weight of phosphoric acid $(H_3PO_4)$ in the reaction $NaOH + H_3PO_4 \to NaH_2PO_4 + H_2O$ is

In terms of energy,$1 \ amu$ is equal to:

Upon fully dissolving $2.0 \ g$ of a metal in sulphuric acid,$6.8 \ g$ of the metal sulphate is formed. The equivalent weight of the metal is $...... \ g$.

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Calculate the average atomic mass of chlorine using the following data:

Isotope Natural abundance (%) Molar mass $(u)$

$^{35}Cl$ $75.77$ $34.9689$

$^{37}Cl$ $24.23$ $36.9659$
(in $u$)