An aqueous solution of urea containing 18 g | vedclass

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Question

(B) Given: Volume of solution $= 1500 \ cm^{3}$,Density of solution $= 1.052 \ g/cm^{3}$,Mass of urea (solute) $= 18 \ g$,Molar mass of urea $= 60 \ g

Vedclass · Accepted Answer

$0.192$

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(B) Given: Volume of solution $= 1500 \ cm^{3}$,Density of solution $= 1.052 \ g/cm^{3}$,Mass of urea (solute) $= 18 \ g$,Molar mass of urea $= 60 \ g/mol$. Mass of solution $= 	ext{Volume} 	imes 	ext{Density} = 1500 \ cm^{3} 	imes 1.052 \ g/cm^{3} = 1578 \ g$. Mass of solvent (water) $= 	ext{Mass of solution} - 	ext{Mass of solute} = 1578 \ g - 18 \ g = 1560 \ g = 1.560 \ kg$. Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in } kg} = \frac{18 / 60}{1.560} = \frac{0.3}{1.560} \approx 0.192 \ m$.

An aqueous solution of urea containing $18 \ g$ of urea in $1500 \ cm^{3}$ of the solution has a density of $1.052 \ g/cm^{3}$. If the molecular weight of urea is $60$,the molality of the solution is: (in $m$)

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