If a 3.6 , M solution of H_2SO_4 contains 29 | vedclass

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Question

(C) Molarity $(M)$ is given by the formula: $M = \frac{	ext{mass percentage} 	imes 	ext{density} 	imes 10}{	ext{Molar mass}}$.Given: $M = 3.6 \,

Vedclass · Accepted Answer

$1.22 \, g \, mL^{-1}$

Vedclass · Answer

(C) Molarity $(M)$ is given by the formula: $M = \frac{	ext{mass percentage} 	imes 	ext{density} 	imes 10}{	ext{Molar mass}}$.Given: $M = 3.6 \, mol \, L^{-1}$,$	ext{mass percentage} = 29 \%$,$	ext{Molar mass} = 98 \, g \, mol^{-1}$.Substituting the values: $3.6 = \frac{29 	imes d 	imes 10}{98}$.$d = \frac{3.6 	imes 98}{290} = \frac{352.8}{290} \approx 1.216 \, g \, mL^{-1}$.Rounding to two decimal places,the density is $1.22 \, g \, mL^{-1}$.

If a $3.6 \, M$ solution of $H_2SO_4$ contains $29 \% \, H_2SO_4$ by mass,find the density of the $H_2SO_4$ solution. (Molar mass of $H_2SO_4 = 98 \, g \, mol^{-1}$)

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