# 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)#?

##### 1 Answer

#### Answer:

#### 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_"urea" = 7.58 * 10^(-2) = 0.0758#

Now, you know that the **total mass** of the solution is equal to

#m_"urea" + m_"water" = "49.0 g"#

with **molar mass** and the number of moles it contains.

#m_"urea" = n_"urea" * M_ "M urea"# Here

#n_"urea"# represents thenumber of molesof urea present in the solution and#M_ "M urea"# is themolar massof urea.

#m_"water" = n_"water" * M_"water"# Here

#n_"water"# represents thenumber of molesof water present in the solution and#M_ "M water"# is themolar massof water.

This means that you can write

# n_"urea" * M_ "M urea" + n_"water" * M_"water" = "49.0 g" " " " "color(darkorange)((1))#

By definition, the mole fraction of urea is

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

which means that you have

#n_"urea"/(n_"urea" + n_"water") = 0.0758" " " " color(darkorange)((2))#

You now have two equations with two unknowns,

#n_"urea" - 0.0758 * n_"urea" = 0.0758 * n_"water"#

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

#n_"water" = 12.19 * n_"urea"#

Plug this into equation

#n_"urea" * M_"M urea" + 12.19 * n_"urea" * M_ "M water" = "49.0 g"#

#n_"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_"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_"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.