Question

Calculate the molarity of the following aqueous solutions: (a) \(0.640 \mathrm{~g}\) of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) in \(500.0 \mathrm{~mL}\) of solution, (b) \(50.0 \mathrm{~g}\) of \(\mathrm{LiClO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) in \(250 \mathrm{~mL}\) of solution, (c) \(125 \mathrm{~mL}\) of \(3.00 \mathrm{M}\) \(\mathrm{HNO}_{3}\) diluted to \(1.00 \mathrm{~L}\)

Short answer

The molarity of the aqueous solutions are: (a) \(0.00862 \, M\) for the \(Mg(NO_3)_2\) solution, (b) \(1.03 \, M\) for the \(LiClO_4 \cdot 3H_2O\) solution, and (c) \(0.375 \, M\) for the diluted \(HNO_3\) solution.

Step-by-step solution

Calculate moles of solute

First, determine the molar mass of \(Mg(NO_3)_2\). The molar mass is the sum of the atomic masses of the elements in the compound. \(Mg(NO_3)_2 = Mg + 2(N + 3O) = 24.31 + 2(14.01+3(16.00))\) Molar mass of \(Mg(NO_3)_2 = 148.33 \, g/mol\) Now, calculate the moles of \(Mg(NO_3)_2\): Moles of \(Mg(NO_3)_2 = \cfrac{0.640 \, g}{148.33 \, g/mol} = 0.00431 \, mol\)

Convert volume to liters

Now, convert the given volume of solution from milliliters to liters: $$500.0 \, mL \times \frac{1 \, L}{1000 \, mL} = 0.500 \, L$$

Calculate molarity

Now, calculate the molarity by dividing moles of solute by the volume of solution in liters: $$M = \cfrac{0.00431 \, mol}{0.500 \, L} = 0.00862 \, M$$ The molarity of the solution is \(0.00862 \, M\). (b) Calculate the molarity of a solution containing 50.0 g of LiClO4·3H2O in 250 mL of solution

Calculate moles of solute

First, determine the molar mass of \(LiClO_4∙3H_2O\): \(LiClO_4∙3H_2O = Li + Cl + 4O + 3(2H + O)\) Molar mass of \(LiClO_4∙3H_2O = 194.10 \, g/mol\) Now, calculate the moles of \(LiClO_4∙3H_2O\): Moles of \(LiClO_4∙3H_2O = \cfrac{50.0 \, g}{194.10 \, g/mol} = 0.257 \, mol\)

Convert volume to liters

Now, convert the given volume of solution from milliliters to liters: $$250.0 \, mL \times \frac{1 \, L}{1000 \, mL} = 0.250 \, L$$

Calculate molarity

Now, calculate the molarity by dividing moles of solute by the volume of solution in liters: $$M = \cfrac{0.257 \, mol}{0.250 \, L} = 1.03 \, M$$ The molarity of the solution is \(1.03 \, M\). (c) Calculate the molarity of 125 mL of a 3.00 M HNO3 solution diluted to 1.00 L

Calculate moles of solute in the initial solution

First, find the moles of \(HNO_3\) in the initial \(125 \, mL\) solution: $$Moles \, of \, HNO_3 = Molarity \times Volume = 3.00 \, M \times 0.125 \, L = 0.375 \, mol$$

Calculate molarity in the diluted solution

The moles of solute do not change during the dilution process, so to calculate the molarity in the final diluted solution, divide the moles of solute by the final volume of solution in liters: $$M = \cfrac{0.375 \, mol}{1.00 \, L} = 0.375 \, M$$ The molarity of the diluted solution is \(0.375 \, M\).

Key concepts

Molar Mass

Molar mass is a key concept when working with chemical solutions and compounds. It's defined as the mass of one mole of a substance, measured in grams per mole (\(g/mol\)). To calculate the molar mass, sum the atomic masses of each element present in the compound according to the chemical formula.
For example, for \(Mg(NO_3)_2\), you would add the atomic masses together:

• Magnesium (Mg): 24.31 g/mol
• Nitrogen (N): 14.01 g/mol, since there are 2 nitrogen atoms, multiply by 2
• Oxygen (O): 16.00 g/mol, and since there are 6 oxygen atoms, multiply by 6

This gives a molar mass for \(Mg(NO_3)_2\) of 148.33 g/mol.
Understanding molar mass allows you to convert between grams and moles, which is crucial in calculating molarity.

Dilution

Dilution involves decreasing the concentration of a solute in a solution, typically by adding more solvent. The main principle of dilution is that the number of moles of solute remains constant, even though the total volume of the solution increases.
To illustrate, when a 125 mL sample of a 3.00 M \(HNO_3\) solution is diluted to a final volume of 1.00 L, the moles of \(HNO_3\) do not change:

• Initial moles of \(HNO_3 = 3.00 \, M \times 0.125 \, L = 0.375 \, mol\)

After dilution, the molarity, or concentration, is calculated by dividing these moles by the new volume:

• \(M = \cfrac{0.375 \, mol}{1.00 \, L} = 0.375 \, M\)

This shows how diluting changes the concentration but not the amount of solute.

Aqueous Solution

An aqueous solution is one where water acts as the solvent. This is one of the most common types of solutions in chemistry. When dealing with aqueous solutions, the solute is dissolved in water, forming a homogeneous mixture.
A key property of aqueous solutions is their capacity to conduct electricity, which occurs if the solute dissolves into ions. For example, \(LiClO_4\) dissolved in water forms an aqueous solution as it dissociates into \(Li^+\) and \(ClO_4^-\) ions.
In molarity calculations, solutions like \(Mg(NO_3)_2\) and \(LiClO_4 \cdot 3H_2O\) are dissolved in water, making them aqueous. It's important to know the volume of the aqueous solution in liters when calculating molarity.

Chemical Formula

Chemical formulas indicate the types and numbers of atoms in a molecule. Understanding a chemical formula allows for calculations like molar mass and informs you about the composition of a compound.
Take \(Mg(NO_3)_2\) as an instance — it shows that one magnesium atom is combined with two nitrate ions. For \(LiClO_4 \cdot 3H_2O\), the formula indicates one lithium perchlorate molecule bound with three water molecules, highlighting the concept of hydrates in chemistry.

• Subscripts denote how many atoms are present.
• Parentheses help group atoms and apply subscripts to entire polyatomic ions.

Recognizing these details from a chemical formula is vital for correctly understanding and solving chemistry problems.