The molality of 15% ,(w/V) solution of H_2SO_ | vedclass

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(D) Given: $15\% \,(w/V)$ solution of $H_2SO_4$ means $15\, g$ of $H_2SO_4$ is present in $100\, mL$ of solution.Density of solution = $1.1\, g/mL$.Ma

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

(D) Given: $15\% \,(w/V)$ solution of $H_2SO_4$ means $15\, g$ of $H_2SO_4$ is present in $100\, mL$ of solution.Density of solution = $1.1\, g/mL$.Mass of solution = $	ext{Volume} 	imes 	ext{Density} = 100\, mL 	imes 1.1\, g/mL = 110\, g$.Mass of solvent = $	ext{Mass of solution} - 	ext{Mass of solute} = 110\, g - 15\, g = 95\, g = 0.095\, kg$.Molar mass of $H_2SO_4 = 98\, g/mol$.Moles of solute = $\frac{15\, g}{98\, g/mol} \approx 0.153\, mol$.Molality $(m) = \frac{	ext{Moles of solute}}{	ext{Mass of solvent in } kg} = \frac{0.153\, mol}{0.095\, kg} \approx 1.61\, mol/kg$.Thus,the correct option is $D$.

The molality of $15\% \,(w/V)$ solution of $H_2SO_4$ with density $1.1\, g/cm^3$ is

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