When 120 g of urea (molar mass = 60 g/mol) | vedclass

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(A) $1$. Calculate the number of moles of urea: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{120 \ g}{60 \ g/mol} = 2 \ mol$. $2$. Calculate the

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$2.05$

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(A) $1$. Calculate the number of moles of urea: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{120 \ g}{60 \ g/mol} = 2 \ mol$. $2$. Calculate the total mass of the solution: $	ext{Mass of solution} = 	ext{mass of solute} + 	ext{mass of solvent} = 120 \ g + 1000 \ g = 1120 \ g$. $3$. Calculate the volume of the solution: $V = \frac{	ext{mass}}{	ext{density}} = \frac{1120 \ g}{1.15 \ g/mL} \approx 973.91 \ mL = 0.97391 \ L$. $4$. Calculate the molarity: $M = \frac{n}{V(L)} = \frac{2 \ mol}{0.97391 \ L} \approx 2.05 \ M$.

When $120 \ g$ of urea (molar mass $= 60 \ g/mol$) is dissolved in $1000 \ g$ of water,the density of the resulting solution is $1.15 \ g/mL$. The molarity of this solution is ........ $M$.

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