The volume of 0.1 M HCl required in mL to n | vedclass

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(C) Given,Molarity of $HCl$ $(M_1) = 0.1 \ M$ Volume of $Na_2CO_3$ solution $(V_2) = 20 \ mL$ Mass of $Na_2CO_3 = 0.106 \ g$ Molar mass of $Na_2CO_3 =

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$20$

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(C) Given,Molarity of $HCl$ $(M_1) = 0.1 \ M$ Volume of $Na_2CO_3$ solution $(V_2) = 20 \ mL$ Mass of $Na_2CO_3 = 0.106 \ g$ Molar mass of $Na_2CO_3 = (2 	imes 23) + 12 + (3 	imes 16) = 106 \ g/mol$ Moles of $Na_2CO_3 = \frac{0.106 \ g}{106 \ g/mol} = 0.001 \ mol$ Molarity of $Na_2CO_3$ $(M_2) = \frac{0.001 \ mol}{0.020 \ L} = 0.05 \ M$ The neutralization reaction is: $Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2$ Using the equivalence relation: $n_{HCl} = 2 	imes n_{Na_2CO_3}$ $M_1 	imes V_1 = 2 	imes (M_2 	imes V_2)$ $0.1 	imes V_1 = 2 	imes (0.05 	imes 20)$ $0.1 	imes V_1 = 2 	imes 1 = 2$ $V_1 = \frac{2}{0.1} = 20 \ mL$

The volume of $0.1 \ M \ HCl$ required in $mL$ to neutralise $20 \ mL$ of a solution containing $0.106 \ g$ of $Na_2CO_3$ is

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