The weight percent of sucrose (formula weight | vedclass

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(B) Given,weight percent of sucrose $= 3.42 \%$. This means $3.42 \, g$ of sucrose is present in $100 \, g$ of solution.Molar mass of sucrose $= 342 \

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$0.1$

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(B) Given,weight percent of sucrose $= 3.42 \%$. This means $3.42 \, g$ of sucrose is present in $100 \, g$ of solution.Molar mass of sucrose $= 342 \, g \, mol^{-1}$.Number of moles of sucrose $= \frac{	ext{mass}}{	ext{molar mass}} = \frac{3.42 \, g}{342 \, g \, mol^{-1}} = 0.01 \, mol$.Density of solution $= 1 \, g \, mL^{-1}$.Volume of solution $= \frac{	ext{mass of solution}}{	ext{density}} = \frac{100 \, g}{1 \, g \, mL^{-1}} = 100 \, mL = 0.1 \, L$.Molarity $= \frac{	ext{moles of solute}}{	ext{volume of solution in } L} = \frac{0.01 \, mol}{0.1 \, L} = 0.1 \, mol \, L^{-1}$.Thus,the correct option is $B$.

The weight percent of sucrose (formula weight $= 342 \, g \, mol^{-1}$) in an aqueous solution is $3.42$. The density of the solution is $1 \, g \, mL^{-1}$. The concentration of sucrose in the solution in $mol \, L^{-1}$ is

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