An aqueous solution containing 6.5 g of NaCl | vedclass

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Question

(C) The reaction for the electrolysis of aqueous $NaCl$ is: $2NaCl + 2H_2O ightarrow 2NaOH + Cl_2 + H_2$. Weight of pure $NaCl = 6.5 \ g 	imes 0.9

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(C) The reaction for the electrolysis of aqueous $NaCl$ is: $2NaCl + 2H_2O ightarrow 2NaOH + Cl_2 + H_2$. Weight of pure $NaCl = 6.5 \ g 	imes 0.9 = 5.85 \ g$. Moles of $NaCl = \frac{5.85 \ g}{58.5 \ g/mol} = 0.1 \ mol$. According to the stoichiometry,$2 \ mol$ of $NaCl$ produces $2 \ mol$ of $NaOH$,so $0.1 \ mol$ of $NaCl$ produces $0.1 \ mol$ of $NaOH$. For neutralization: $NaOH + CH_3COOH ightarrow CH_3COONa + H_2O$. Moles of $CH_3COOH$ required = Moles of $NaOH = 0.1 \ mol$. Volume of $1 \ M$ acetic acid = $\frac{	ext{moles}}{	ext{molarity}} = \frac{0.1 \ mol}{1 \ M} = 0.1 \ L = 100 \ cm^{3}$.

Column $I$	Column $II$
$(A)$. Kohlrausch law can calculate	$(P)$. $\frac{\Lambda_m^c}{\Lambda_m^o}$
$(B)$. Molar conductance $\Lambda_m$	$(Q)$. $\frac{1}{R} \times \frac{l}{A}$
$(C)$. Specific conductance $\kappa$	$(R)$. $\Lambda_m^o$ of $Ca_3(PO_4)_2$
$(D)$. Degree of ionization of weak electrolyte	$(S)$. $\frac{\kappa \times 1000}{M}$

An aqueous solution containing $6.5 \ g$ of $NaCl$ of $90 \%$ purity was subjected to electrolysis. After the complete electrolysis,the solution was evaporated to get solid $NaOH$. The volume of $1 \ M$ acetic acid required to neutralize $NaOH$ obtained above is (in $cm^{3}$)

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