# A solution of phosphoric acid was made by dissolving 10.0 g #H_3PO_4# in 100.0 mL water. The resulting volume was 104 mL. What is the density, mole fraction, molarity, and molality of the solution?

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Assume water has a density of 1.00 #g# #/# #cm^3# .

Assume water has a density of 1.00

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The first thing to focus on here is finding the **density** of the solution.

As you know, density is defined as mass per unit of volume. You know that your solution has a volume of

Notice that you are given the **volume** of water and its density. Use this information to find the **mass of water** used to make the solution - keep in mind that you have

#100.0 color(red)(cancel(color(black)("mL"))) * overbrace("1.00 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("density of water")) = "100.0 g"#

So, the **total mass** of the solution will be equal to

#m_"sol" = m_(H_3PO_4) + m_"water"#

#m_"sol" = "10.0 g" + "100.0 g" = "110.0 g"#

The density of the solution will thus be

#rho = "110.0 g"/"104 mL" = color(green)("1.06 g mL"^(-1))#

To get the **mole fraction** of phosphoric acid in this solution, you need to know

thenumber of molesof phosphoric acidthenumber of molesof water

Since you have the masses of the two compounds, you can use their **molar masses** to determine how many moles of each you have present in this solution.

#10.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_3"PO"_4)/(97.995color(red)(cancel(color(black)("g")))) = "0.10205 moles H"_3"PO"_4#

and

#100.0 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "5.551 moles H"_2"O"#

The **total number of moles** in this solution will be

#n_"total" = "0.1025 moles" + "5.551 moles" = "5.6531 moles"#

The mole fraction of phosphoric acid will be

#chi_(H_3PO_4) = (0.10205 color(red)(cancel(color(black)("moles"))))/(5.6531 color(red)(cancel(color(black)("moles")))) = color(green)(0.0181)#

The **molarity** of the solution is defined as the number of moles of solute divided by the volume of the solution - expressed in **liters**.

#color(blue)(c = n_"solute"/V_"solution")#

In your case, you will have

#c = "0.10205 moles"/(104 * 10^(-3)"L") = color(green)("0.981 mol L"^(-1))#

Finally, the **molality** of the solution is defined as the number of moles of solute divided by the *mass of the solvent* - expressed in **kilograms**.

#color(blue)(b = n_"solute"/m_"solvent")#

Plug in your values to get

#b = "0.10205 moles"/(100.0 * 10^(-3)"kg") = color(green)("1.02 mol kg"^(-1))#