30.0 mL of the given HCl solution requires 2 | vedclass

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(C) Given:Volume of $HCl$ solution $(V_1) = 30 \ mL$Volume of sodium carbonate solution $(Na_2CO_3)$ $(V_2) = 20 \ mL$Concentration of $Na_2CO_3$ solu

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) Given:Volume of $HCl$ solution $(V_1) = 30 \ mL$Volume of sodium carbonate solution $(Na_2CO_3)$ $(V_2) = 20 \ mL$Concentration of $Na_2CO_3$ solution $(M_2) = 0.1 \ M$Volume of $NaOH$ solution $(V_3) = 30.0 \ mL$Concentration of $NaOH$ solution $(M_3) = 0.2 \ M$For the reaction between $HCl$ and $Na_2CO_3$ $(Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2)$:$n_{HCl} = 2 	imes n_{Na_2CO_3}$$M_1V_1 = 2 	imes M_2V_2$$M_1 = \frac{2 	imes M_2 	imes V_2}{V_1} = \frac{2 	imes 0.1 	imes 20}{30} = \frac{4}{30} \ M$Also,for the reaction between $HCl$ and $NaOH$ $(NaOH + HCl ightarrow NaCl + H_2O)$:$M_1 V_f = M_3 V_3$ (where $V_f$ is the volume of $HCl$ solution required)$V_f = \frac{M_3 	imes V_3}{M_1} = \frac{0.2 	imes 30}{4 / 30} = \frac{6}{4 / 30} = \frac{180}{4} = 90 \ mL$Hence,option $C$ is the correct answer.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

$30.0 \ mL$ of the given $HCl$ solution requires $20.0 \ mL$ of $0.1 \ M$ sodium carbonate solution for complete neutralisation. What is the volume of this $HCl$ solution required to neutralise $30.0 \ mL$ of $0.2 \ M$ $NaOH$ solution (in $mL$)?

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