What is the molarity of a 1 L solution of 93 | vedclass

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(A) Given: $93\% \ (w/v)$ $H_2SO_4$ means $93 \ g$ of $H_2SO_4$ is present in $100 \ mL$ of solution.Therefore,in $1000 \ mL$ $(1 \ L)$ of solution,th

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$10.43$

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(A) Given: $93\% \ (w/v)$ $H_2SO_4$ means $93 \ g$ of $H_2SO_4$ is present in $100 \ mL$ of solution.Therefore,in $1000 \ mL$ $(1 \ L)$ of solution,the mass of $H_2SO_4$ is $930 \ g$.Molar mass of $H_2SO_4 = 2 	imes 1 + 32 + 4 	imes 16 = 98 \ g/mol$.Number of moles of $H_2SO_4 = \frac{	ext{Mass}}{	ext{Molar mass}} = \frac{930}{98} \approx 9.49 \ mol$.Molarity $(M) = \frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{9.49}{1} = 9.49 \ M$.Note: The provided options suggest the question might be asking for molality or there is a discrepancy in the provided options. Based on standard molarity calculation,the result is $9.49 \ M$. However,if we calculate molality $(m)$ using density $d = 1.84 \ g/mL$:Mass of solution $= 1000 \ mL 	imes 1.84 \ g/mL = 1840 \ g$.Mass of solvent $= 1840 \ g - 930 \ g = 910 \ g = 0.91 \ kg$.Molality $(m) = \frac{9.49 \ mol}{0.91 \ kg} \approx 10.43 \ m$. Thus,the correct answer corresponds to the calculated molality.

What is the molarity of a $1 \ L$ solution of $93\% \ (w/v)$ $H_2SO_4$ (in $M$)? The density of the solution is $1.84 \ g/mL$.

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