What is the density (in g mL^{-1}) of a 29% | vedclass

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(C) The relationship between molarity $(M)$,density ($d$ in $g \ mL^{-1}$),and percentage by mass $(w/w\%)$ is given by the formula: $M = \frac{w/w\%

Vedclass · Accepted Answer

$1.22$

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(C) The relationship between molarity $(M)$,density ($d$ in $g \ mL^{-1}$),and percentage by mass $(w/w\%)$ is given by the formula: $M = \frac{w/w\% 	imes d 	imes 10}{Molar \ mass}$ Given: $M = 3.60 \ M$ $w/w\% = 29\%$ $Molar \ mass = 98 \ g \ mol^{-1}$ Substituting the values: $3.60 = \frac{29 	imes d 	imes 10}{98}$ $3.60 = \frac{290 	imes d}{98}$ $d = \frac{3.60 	imes 98}{290}$ $d = \frac{352.8}{290} \approx 1.2165 \ g \ mL^{-1}$ Rounding to two decimal places,we get $d \approx 1.22 \ g \ mL^{-1}$.

What is the density (in $g \ mL^{-1}$) of a $29\%$ (by mass) $H_2SO_4$ solution that has a molarity of $3.60 \ M$? (Molar mass of $H_2SO_4 = 98 \ g \ mol^{-1}$)

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