# Find the mass of urea, "CO"("NH"_2)_2, needed to prepare "49.0 g" of a solution in water in which the mole fraction of urea is 7.58 * 10^(−2)?

Feb 9, 2018

$1.05 \cdot {10}^{1}$ $\text{g}$

#### Explanation:

The mole fraction of urea is defined as the ratio between the number of moles of urea and the total number of moles present in the solution.

In your case, the mole fraction of urea is said to be equal to

${\chi}_{\text{urea}} = 7.58 \cdot {10}^{- 2} = 0.0758$

Now, you know that the total mass of the solution is equal to $\text{49.0 g}$. You can write this as

${m}_{\text{urea" + m_"water" = "49.0 g}}$

with ${m}_{\text{urea}}$ being the mass of urea and ${m}_{\text{water}}$ being the mass of water. As you know, you can express the mass of a substance using its molar mass and the number of moles it contains.

${m}_{\text{urea" = n_"urea" * M_ "M urea}}$

Here ${n}_{\text{urea}}$ represents the number of moles of urea present in the solution and ${M}_{\text{M urea}}$ is the molar mass of urea.

${m}_{\text{water" = n_"water" * M_"water}}$

Here ${n}_{\text{water}}$ represents the number of moles of water present in the solution and ${M}_{\text{M water}}$ is the molar mass of water.

This means that you can write

${n}_{\text{urea" * M_ "M urea" + n_"water" * M_"water" = "49.0 g" " " " }} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$

By definition, the mole fraction of urea is

chi_"urea" = n_"urea"/(n_"urea" + n_"water")

which means that you have

${n}_{\text{urea"/(n_"urea" + n_"water") = 0.0758" " " }} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(2\right)}$

You now have two equations with two unknowns, ${n}_{\text{urea}}$ and ${n}_{\text{water}}$. Use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(2\right)}$ to say that

${n}_{\text{urea" - 0.0758 * n_"urea" = 0.0758 * n_"water}}$

n_"water" = (0.9242 * n_"urea")/0.0758

${n}_{\text{water" = 12.19 * n_"urea}}$

Plug this into equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$ to find

${n}_{\text{urea" * M_"M urea" + 12.19 * n_"urea" * M_ "M water" = "49.0 g}}$

${n}_{\text{urea" * (M_ "M urea" + 12.19 * M_ "M water") = "49.0 g}}$

This will get you

n_"urea" = "49.0 g"/(M_ "M urea" + 12.19 * M_ "M water")

Plug in the molar mass of urea and the molar mass of water to get

n_"urea" = (49.0 color(red)(cancel(color(black)("g"))))/((60.06 + 12.19 * 18.015) color(red)(cancel(color(black)("g"))) "mol"^(-1))

${n}_{\text{urea" = "0.1752 moles}}$

Finally, to find the mass of urea, use the molar mass of the compound.

0.1752 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = "10.5 g"

Therefore, you can say that this solution contains

${m}_{\text{urea" = color(darkgreen)(ul(color(black)(1.05 * 10^1 quad "g}}$

of urea. The answer is rounded to three sig figs and expressed in scientific notation.