The density (in g ; mL^{-1}) of a 3.60 ; M su | vedclass

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Question

(D) The relationship between Molarity $(M)$,percentage by mass $(\% \; w/w)$,and density $(d)$ is given by the formula: $M = \frac{10 	imes \% \; w/w

Vedclass · Accepted Answer

$1.22$

Vedclass · Answer

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

The density (in $g \; mL^{-1}$) of a $3.60 \; M$ sulphuric acid solution that is $29 \%$ $H_2SO_4$ (molar mass $= 98 \; g \; mol^{-1}$) by mass will be

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