# a) How many moles of Al_2(SO_4)_3 are required to make 45 mL of a 0.080 M Al_2(SO_4)_3 solution?...

## Question:

a) How many moles of {eq}Al_2(SO_4)_3 {/eq} are required to make 45 mL of a 0.080 M {eq}Al_2(SO_4)_3 {/eq} solution? (Hint, molarity is moles per liter. You know the molarity. You are given the number of mL's, which can be converted to liters. Just set it up so the units cancel and you get "moles".)

b) What is the molar mass of {eq}Al_2(SO_4)_3 {/eq}, to the nearest gram?

c) What mass of {eq}Al_2(SO_4)_3 {/eq} is required to make 45 mL of a 0.080 M {eq}Al_2(SO_4)_3 {/eq} solution?

(If you've got moles and the molecular weight you can get to grams.)

d) How many moles of aluminum ions are present in the solution?

## Molar mass, molarity, and molecular formula:

Molarity is the concentration of a solution expressed as moles of solute per liter of solution. Molar mass is the mass of all the atoms in a molecule expressed in grams per mole. To calculate the molar mass, obtain the sum of the atomic weights of the individual elements each multiplied to the number of times the element appeared in the molecule. For example, the molar mass of {eq}H_2O{/eq} is equal to the twice the atomic weight of H plus the atomic weight of O.

{eq}MM_{H_2O} = (A_H \times 2 \ atoms) + (A_O \times 1 \ atom) {/eq} The molecular formula of a compound indicates which atoms and how many of those atoms are present in the molecule.

## Answer and Explanation:

a. Molarity is defined as moles solute per liter solution, i.e.

{eq}Molarity = \dfrac{moles}{liter \ solution}{/eq}

The number of moles {eq}Al_2(SO_4)_3{/eq} can be obtained by rearranging the equation and plugging in the known values.

{eq}\begin{align*} moles \ Al_2(SO_4)_3 &= Molarity \times Volume solution (in \ L)\\ &= (0.080M)(0.045L)\\ &= \boxed{0.0036 \ moles} \end{align*} \\ \\ {/eq}

b. List down the atomic weight of the elements making up {eq}Al_2(SO_4)3{/eq}:

{eq}Al: 26.982 \ amu\\ S: 32.066 \ amu\\ O: 15.999 \ amu\\ \\ {/eq}

The molar mass of {eq}\begin{align*} MM_{Al_2(SO_4)3} &= (A_{Al} \times 2) + (A_S \times 3) + (A_O \times 12) \\ &= (26.982 amu \times 2) + (32.066 \times 3) + (15.999 \times 12) \\ &= \boxed{342.15\dfrac{g}{mol}}\\ \end{align*} \\ \\ {/eq}

c. From (a) we know that 0.0036 moles of {eq}{Al_2(SO_4)3}{/eq} is needed to make 45 ml of 0.080 M {eq}Al_2(SO_4)_3{/eq}. Converting moles to mass using the molar mass of {eq}Al_2(SO_4)_3{/eq}:

{eq}\begin{align*} mass \ Al_2(SO_4)_3 &= moles \ Al_2(SO_4)_3 \times MM \ Al_2(SO_4)_3\\ &= 0.0036 \ moles \ Al_2(SO_4)_3 \times \dfrac{342.15 \ g}{mol}\\ &= \boxed{1.23 \ g} \\ \end{align*} \\ \\ {/eq}

d. From the molecular formula {eq}Al_2(SO_4)_3{/eq}, we know that there are 2 moles of aluminum ions for every mole of aluminum sulfate. Using this relation, the moles of aluminum ions can be obtained.

{eq}\begin{align*} moles \ Al^{3+} &= 0.0036 \ moles \ Al_2(SO_4)_3 \times \dfrac{2 \ moles \ Al^{3+}}{1 \ mole \ Al_2(SO_4)_3}\\ &= \boxed{0.0072 \ moles}\\ \end{align*} {/eq}

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Chapter 27 / Lesson 24