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(C) Total hardness $= 200 \ ppm$ of $Mg^{2+}$.Temporary hardness is due to $Mg(HCO_3)_2$ and permanent hardness is due to $MgCl_2$.$Na_2CO_3$ removes

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$1.9$

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(C) Total hardness $= 200 \ ppm$ of $Mg^{2+}$.Temporary hardness is due to $Mg(HCO_3)_2$ and permanent hardness is due to $MgCl_2$.$Na_2CO_3$ removes temporary hardness by precipitating $Mg^{2+}$ as $MgCO_3$.Remaining hardness is due to $MgCl_2$,which is $\frac{1000}{3} \ ppm$ of $CaCO_3$ equivalent.$1 \ ppm = 1 \ mg/L$. So,remaining hardness $= \frac{1000}{3} \ mg/L$ of $CaCO_3$.In $6 \ L$,total $CaCO_3$ equivalent $= \frac{1000}{3} 	imes 6 = 2000 \ mg = 2 \ g$ of $CaCO_3$.Moles of $CaCO_3 = \frac{2 \ g}{100 \ g/mol} = 0.02 \ mol$.Since $1 \ mol$ of $MgCl_2$ is equivalent to $1 \ mol$ of $CaCO_3$,moles of $MgCl_2 = 0.02 \ mol$.Mass of $MgCl_2 = 0.02 \ mol 	imes 95 \ g/mol = 1.9 \ g$.

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

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