Question

Calculate the number of moles of solute present in each of the following solutions: (a) 255 \(\mathrm{mL}\) of \(1.50 \mathrm{MNO}_{3}(a q),\) (b) 50.0 \(\mathrm{mg}\) of an aqueous solution that is 1.50 \(\mathrm{m} \mathrm{NaCl}_{1}\) (c) 75.0 g of an aqueous solution that is 1.50\(\%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.

Short answer

In summary: (a) 0.3825 moles of MNO3 are present in the 255 mL of 1.50 M solution. (b) 0.000075 moles of NaCl are present in the 50.0 mg of 1.50 m solution. (c) 0.00329 moles of sucrose are present in the 75.0 g of 1.50% solution.

Step-by-step solution

(a) Solution Volume and Molarity

Given the volume (255 mL) and the molarity (1.50 M) of the MNO3 solution, we can calculate the number of moles of solute using the formula: Moles of solute = Molarity × Volume (in L) First, convert the volume from mL to L: \(255 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.255 \mathrm{L}\) Now, use the formula: Moles of MNO3 = 1.50 M × 0.255 L = 0.3825 moles

(b) Solution Mass and Molality

Given the mass of the solution (50.0 mg) and the molality (1.50 m) of the NaCl solution, we can calculate the number of moles of solute using the formula: Moles of solute = Molality × Mass of solution (in kg) First, convert the mass of the solution from mg to kg: \(50.0 \mathrm{mg} \times \frac{1 \mathrm{g}}{1000 \mathrm{mg}} \times \frac{1 \mathrm{kg}}{1000 \mathrm{g}} = 0.00005 \mathrm{kg}\) Now, use the formula: Moles of NaCl = 1.50 m × 0.00005 kg = 0.000075 moles

(c) Solution Mass and Mass Percent

Given the mass of the solution (75.0 g) and the mass percent (1.50%) of the sucrose (\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)) solution, we can calculate the mass of sucrose and then determine the number of moles using its molar mass. First, find the mass of sucrose in the solution: Mass of sucrose = Mass percent × Total mass of solution \(Mass \, of \, sucrose = 1.50\% \times 75.0 \, g = 0.015 \times 75.0 \, g = 1.125 \, g\) Next, determine the molar mass of sucrose: Molar mass of sucrose = \(12 \times 12.01 + 22 \times 1.01 + 11 \times 16.00 = 342.30 \, g/mol\) Then, use the mass of sucrose and the molar mass to find the moles of sucrose: Moles of sucrose = \(\frac{Mass \, of \, sucrose}{Molar \, mass \, of \, sucrose}\) Moles of sucrose = \(\frac{1.125 \, g}{342.30 \, g/mol} = 0.00329 \, moles\) In summary: (a) 0.3825 moles of MNO3 are present in the solution. (b) 0.000075 moles of NaCl are present in the solution. (c) 0.00329 moles of sucrose are present in the solution.

Key concepts

Molarity and Volume Relationship

Understanding the relationship between molarity and volume is crucial for students learning to calculate the number of moles of solute in a solution. Molarity (M) is a measure of concentration that expresses the moles of a solute per liter of solution. Therefore, if you're given the volume of the solution in liters and its molarity, you can calculate the moles of solute present with a simple multiplication.

The formula to find the number of moles (Moles of solute = Molarity \(\times\) Volume) requires that the volume is in liters, not milliliters or any other unit. This is a common source of mistakes among students. When dealing with problems, make sure to first convert the volume to liters, which is done by dividing the number of milliliters by 1000, since there are 1000 milliliters in one liter.Here's a practical example from the exercise given:Firstly, the volume of the solution was converted into liters (255 \mathrm{mL} = 0.255 \mathrm{L}). With the molarity stated as 1.50 M (which reads as '1.50 moles per liter'), finding the moles of MNO_{3} became a straightforward calculation:\(Moles of MNO3 = 1.50 M \times 0.255 L = 0.3825 moles\).Always pay special attention to units when making these calculations to avoid any errors and ensure that the result is in the correct form.

Molality and Solution Mass Relationship

Molality (m), not to be confused with molarity, is another measure of concentration. It is defined as the moles of solute per kilogram of solvent. This relationship is particularly useful when dealing with changes in temperature or pressure, as molality doesn't change with such conditions unlike molarity. The molality is therefore critical in certain conditions and applications, especially in boiling point elevation and freezing point depression calculations.

To compute the moles of solute using molality, the mass of the solution needs to be converted to kilograms, as molality is given in moles per kilogram. You can do this by first converting the mass in milligrams to grams, then grams to kilograms (1 kg = 1000 g = 1,000,000 mg).Considering the example from our textbook:The mass of the NaCl solution was converted into kilograms (50.0 \mathrm{mg} = 0.00005 \mathrm{kg}), and the molality was given as 1.50 m. Using the formula (Moles of solute = Molality \times Mass of solution (in kg)), we find the number of moles of NaCl:\(Moles of NaCl = 1.50 m \times 0.00005 kg = 0.000075 moles\).Remember that the molality involves solvent mass only, not the total solution mass, which includes both the solute and the solvent.

Mass Percent and Solution Mass Relationship

Mass percent concentration is another approach to express the concentration of a solution. It signifies the mass of the solute divided by the total mass of the solution, multiplied by 100%. This percentage tells us how many grams of solute are in every 100 grams of solution. Understanding how to convert mass percent into the actual mass of solute in grams is important to calculate the moles of solute present.

Once you know the mass percent and the total solution mass, you can find the mass of the solute by using the following formula:\(Mass of solute = Mass percent \times Total mass of solution\).For the sucrose example from our exercise, the mass of soluce can be found as follows:\(Mass of sucrose = 1.50\% \times 75.0 g = 0.015 \times 75.0 g = 1.125 g\).After determining the mass of the solute, use the molar mass of the solute to find out the number of moles. The molar mass is the mass of one mole of any chemical compound, such as sucrose here. To find the number of moles, divide the mass of the solute by its molar mass:\(Moles of sucrose = \frac{Mass of sucrose}{Molar mass of sucrose}\).Using the molar mass of sucrose (342.30 g/mol), the number of moles of sucrose in the solution is calculated as 0.00329 moles. This process is essential to understand when working with solutions and analyzing their concentrations.