A solution of H_2SO_4 is 31.4 % H_2SO_4 by ma | vedclass

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(A) The mass percentage of $H_2SO_4$ is $31.4 \%$,which means $31.4 \ g$ of $H_2SO_4$ is present in $100 \ g$ of the solution.The volume of $100 \ g$

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$4$

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(A) The mass percentage of $H_2SO_4$ is $31.4 \%$,which means $31.4 \ g$ of $H_2SO_4$ is present in $100 \ g$ of the solution.The volume of $100 \ g$ of the solution is calculated using density $(d = 1.25 \ g \ mL^{-1})$:$V = \frac{	ext{mass}}{	ext{density}} = \frac{100 \ g}{1.25 \ g \ mL^{-1}} = 80 \ mL$.The number of moles of $H_2SO_4$ is:$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{31.4 \ g}{98 \ g \ mol^{-1}} \approx 0.3204 \ mol$.Molarity $(M)$ is defined as moles of solute per liter of solution:$M = \frac{n}{V(L)} = \frac{0.3204 \ mol}{0.080 \ L} = 4.005 \ M$.Rounding to the nearest integer,the molarity is $4 \ M$.

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

$A$ solution of $H_2SO_4$ is $31.4 \%$ $H_2SO_4$ by mass and has a density of $1.25 \ g \ mL^{-1}$. The molarity of the $H_2SO_4$ solution is . . . . . . $M$ (nearest integer) [Given molar mass of $H_2SO_4 = 98 \ g \ mol^{-1}$]

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