If 0.0100 g of urea [(NH_2)_2CO] is added to | vedclass

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(B) Given: Mass of urea $= 0.0100 \ g$. Molecular mass of urea $[(NH_2)_2CO] = 60 \ g/mol$. Number of moles of urea $= \frac{0.0100 \ g}{60 \ g/mol} =

Vedclass · Accepted Answer

$5.55 	imes 10^{-4} \ m$

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(B) Given: Mass of urea $= 0.0100 \ g$. Molecular mass of urea $[(NH_2)_2CO] = 60 \ g/mol$. Number of moles of urea $= \frac{0.0100 \ g}{60 \ g/mol} = 1.667 	imes 10^{-4} \ mol$. Volume of water $= 0.3000 \ m^3 = 300 \ L = 300,000 \ mL$. Since density of water is $1 \ g/mL$,mass of water $= 300,000 \ g = 300 \ kg$. Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{1.667 	imes 10^{-4} \ mol}{300 \ kg} = 5.55 	imes 10^{-7} \ m$. Note: Based on the provided options,the calculation assumes the volume was $0.3000 \ L$ instead of $0.3000 \ m^3$. If volume $= 0.3000 \ L$,then mass $= 0.3 \ kg$,and $m = \frac{1.667 	imes 10^{-4}}{0.3} = 5.55 	imes 10^{-4} \ m$.

If $0.0100 \ g$ of urea $[(NH_2)_2CO]$ is added to $0.3000 \ m^3$ of water,what is the molality of the urea solution?

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