Problem 27
Question
What is the molality of a solution containing 10.0 \(\mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) dissolved in \(1000,0\) g of water?
Step-by-Step Solution
Verified Answer
The molality of the solution is 0.0704 mol/kg.
1Step 1: Calculate the Molar Mass of the Solute
The first step is to compute the molar mass of sodium sulfate, \( \mathrm{Na}_2\mathrm{SO}_4 \). Using atomic masses: \( \mathrm{Na} = 22.99 \; \mathrm{g/mol} \), \( \mathrm{S} = 32.07 \; \mathrm{g/mol} \), and \( \mathrm{O} = 16.00 \; \mathrm{g/mol} \), the molar mass is computed as follows: \[ \text{Molar mass of } \mathrm{Na}_2\mathrm{SO}_4 = 2(22.99) + 32.07 + 4(16.00) = 142.05 \; \mathrm{g/mol} \]
2Step 2: Determine the Moles of Solute
With the molar mass known, convert 10.0 g of \( \mathrm{Na}_2\mathrm{SO}_4 \) to moles using the molar mass: \[ \text{Moles of } \mathrm{Na}_2\mathrm{SO}_4 = \frac{10.0 \; \mathrm{g}}{142.05 \; \mathrm{g/mol}} \approx 0.0704 \; \mathrm{mol} \]
3Step 3: Convert the Mass of Solvent to Kilograms
Molality is expressed in moles of solute per kilogram of solvent. Here, water is the solvent, and its mass is given as 1000.0 g. Convert grams to kilograms:\[ 1000.0 \; \mathrm{g} = 1.0000 \; \mathrm{kg} \]
4Step 4: Calculate the Molality of the Solution
Molality (\( m \)) is calculated using the formula \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). Substitute the known values:\[ m = \frac{0.0704 \; \mathrm{mol}}{1.0000 \; \mathrm{kg}} = 0.0704 \; \mathrm{mol/kg} \]
Key Concepts
Molar Mass CalculationConversion of Grams to KilogramsMoles of Solute
Molar Mass Calculation
When calculating the molar mass of a compound, it's essential to understand what molar mass is and why it's important. Molar mass represents the mass of one mole of a substance, expressed in grams per mole (g/mol). This value helps convert between the mass of a compound and the number of moles, a key step in stoichiometric calculations.
To find the molar mass, sum up the atomic masses of all atoms in the formula of the compound. Let's look at sodium sulfate (\( \mathrm{Na}_2\mathrm{SO}_4 \)) as our example:
To find the molar mass, sum up the atomic masses of all atoms in the formula of the compound. Let's look at sodium sulfate (\( \mathrm{Na}_2\mathrm{SO}_4 \)) as our example:
- Two sodium (\( \mathrm{Na} \)) atoms, with an atomic mass of about 22.99 g/mol each, amount to\( 2 imes 22.99 = 45.98 \) g/mol.
- One sulfur (\( \mathrm{S} \)) atom with an atomic mass of 32.07 g/mol adds to the total.
- Four oxygen (\( \mathrm{O} \)) atoms each having an atomic mass of 16.00 g/mol, contribute \( 4 imes 16.00 = 64.00 \) g/mol.
Conversion of Grams to Kilograms
Understanding how to convert grams to kilograms is crucial especially when calculating properties of a solution, like molality. Grams and kilograms are both units of mass; however, kilograms are a larger unit commonly used for bigger quantities.
- 1 kilogram is equivalent to 1000 grams. This means when you're given a mass in grams, you can convert it to kilograms by dividing the number by 1000.
Moles of Solute
Calculating the moles of a solute is an essential step in determining concentrations in a solution. Moles relate a given mass of a chemical substance to the number of particles or molecules it contains. The key to finding the number of moles is the formula:\[\text{Moles} = \frac{\text{mass of solute (in g)}}{\text{molar mass of solute (in g/mol)}}\]Let's take sodium sulfate (\( \mathrm{Na}_2\mathrm{SO}_4 \)) as an example. Suppose you have 10.0 grams of this compound. Given its molar mass is 142.05 g/mol, the moles of sodium sulfate are:\[\text{Moles of } \mathrm{Na}_2\mathrm{SO}_4 = \frac{10.0 \, \mathrm{g}}{142.05 \, \mathrm{g/mol}} \approx 0.0704 \, \mathrm{mol}\]Understanding how to convert grams to moles allows you to express chemical quantities in terms of a number that represents how much substance you have, which is crucial for further chemical calculations, like finding the molality of solutions.
Other exercises in this chapter
Problem 25
How many milliliters of a 5.0\(M \mathrm{H}_{2} \mathrm{SO}_{4}\) stock solution would you need to prepare 100.0 \(\mathrm{mL}\) of 0.25\(M \mathrm{H}_{2} \math
View solution Problem 26
Challenge If 0.5 \(\mathrm{L}\) of 5 \(\mathrm{M}\) stock solution of HCl is diluted to make 2 \(\mathrm{L}\) of solution, how much \(\mathrm{HCl},\) in grams,
View solution Problem 29
What is the mole fraction of \(\mathrm{NaOH}\) in an aqueous solution that contains 22.8\(\% \mathrm{NaOH}\) by mass?
View solution Problem 32
Explain the similarities and differences between a 1\(M\) solution of \(\mathrm{NaOH}\) and a 1 \(\mathrm{m}\) solution of \(\mathrm{NaOH} .\)
View solution