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\( 25 \mathrm{~cm}^{3} \) of oxalic acid completely neutralized \( 0.064 \mathrm{~g} \) of sodium hydroxide. Molarity of the
oxalic acid solution is
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oxalic acid solution is
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The correct answer is:
\( 0.032 \)
Given,
$V$ (oxalic acid) $=25 \mathrm{~cm}^{3}$
$W(N a O H)=0.064 g$
$M$ (oxalic acid) $=?$
Since, moles of oxalic acid $=$ mole of $N a O H$
$V$ (oxalic acid) $\times N$ (oxalic acid) $=\frac{w}{M}(N a O H)$
$\Rightarrow N$ ( oxalic acid) $=\frac{0.064 \times 1000}{25 \times 40}=0.064$
Normality $=$ Molarity $\times$ Basicity
$\Rightarrow$ Molarity $=\frac{0.064}{2}=0.32$
Note: Basicity of oxalic acid is 2
$V$ (oxalic acid) $=25 \mathrm{~cm}^{3}$
$W(N a O H)=0.064 g$
$M$ (oxalic acid) $=?$
Since, moles of oxalic acid $=$ mole of $N a O H$
$V$ (oxalic acid) $\times N$ (oxalic acid) $=\frac{w}{M}(N a O H)$
$\Rightarrow N$ ( oxalic acid) $=\frac{0.064 \times 1000}{25 \times 40}=0.064$
Normality $=$ Molarity $\times$ Basicity
$\Rightarrow$ Molarity $=\frac{0.064}{2}=0.32$
Note: Basicity of oxalic acid is 2
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