A certain aqueous solution of FeCl_3 (formula | vedclass

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(D) $1$. Given: Density of solution $(d)$ $= 1.1 \ g/mL$,Mass percentage of $FeCl_3 = 20.0 \ \%$,Molar mass of $FeCl_3 = 162 \ g/mol$. $2$. Consider $

Vedclass · Accepted Answer

$1.47$

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(D) $1$. Given: Density of solution $(d)$ $= 1.1 \ g/mL$,Mass percentage of $FeCl_3 = 20.0 \ \%$,Molar mass of $FeCl_3 = 162 \ g/mol$. $2$. Consider $100 \ g$ of the solution. $3$. Mass of $FeCl_3 = 20.0 \ g$. $4$. Number of moles of $FeCl_3 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{20.0 \ g}{162 \ g/mol} \approx 0.1235 \ mol$. $5$. Volume of $100 \ g$ solution $= \frac{	ext{mass}}{	ext{density}} = \frac{100 \ g}{1.1 \ g/mL} \approx 90.91 \ mL = 0.09091 \ L$. $6$. Molarity $(M)$ $= \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.1235 \ mol}{0.09091 \ L} \approx 1.358 \ M$. $7$. Re-evaluating the calculation: $M = \frac{\% 	imes d 	imes 10}{M_w} = \frac{20.0 	imes 1.1 	imes 10}{162} = \frac{220}{162} \approx 1.358 \ M$. $8$. Since the closest option provided is $1.36 \ M$ (approx),but given the options,there might be a slight discrepancy in the provided choices. However,calculating $1.36 \ M$ is the standard result. Given the options,$1.36 \ M$ is closest to $1.27 \ M$ or $1.47 \ M$. Re-checking the math: $220 / 162 = 1.358$. The correct value is $1.36 \ M$.

$A$ certain aqueous solution of $FeCl_3$ (formula mass $= 162 \ g/mol$) has a density of $1.1 \ g/mL$ and contains $20.0 \ \% \ FeCl_3$ by mass. The molar concentration (molarity) of this solution is: (in $M$)

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