At 25 °C, the density of a 60.0% (m/m) solution of sucrose, C12H22O11, in water is 1.29 g/mL. What is the molarity of this solution?
Unfortuantly the meaning of m/m has two meanings, mass solute/total mass, or mass solute/masssolvent.
I will work it with the first usage.
A liter of solution will have a mass of 1.29kg, and in that, sucrose will be .6*1.29=774 grams sugar.
Molarity=(774/342)/1= you do it.
Now, if your text, or prof uses the other common definition of m/m, calculate that.
To find the molarity of the solution, we need to first convert the given density to grams of sucrose per liter of solution.
1. Convert the given density from g/mL to g/L:
Density = 1.29 g/mL
To convert g/mL to g/L, we multiply by 1000 since there are 1000 mL in 1 L:
Density = 1.29 g/mL * 1000 mL/L
= 1290 g/L
2. Calculate the mass of sucrose in 1 L of the solution:
Since the solution is 60.0% (m/m) sucrose, we can assume that 1000 g of the solution contains 60% sucrose.
Mass of sucrose in 1000 g of solution = 60% of 1000 g
= (60/100) * 1000 g
= 600 g
Therefore, there are 600 g of sucrose in 1 L of the solution.
3. Calculate the molarity of the solution:
Molarity (M) is defined as moles of solute per liter of solution.
To find the moles of sucrose, we need to know its molar mass.
The molar mass of sucrose (C12H22O11) can be calculated by summing the atomic masses:
Molar mass of C12H22O11 = 12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol (approximately)
Now, we can calculate the molarity:
Molarity = Moles of sucrose / Volume of solution (in L)
Moles of sucrose = Mass of sucrose / Molar mass of sucrose
= 600 g / 342.34 g/mol (approximately)
Finally, divide the moles of sucrose by the volume of the solution in liters:
Molarity = (600 g / 342.34 g/mol) / 1 L
≈ 1.75 mol/L
Therefore, the molarity of the 60.0% (m/m) solution of sucrose in water at 25 °C is approximately 1.75 mol/L.
To find the molarity (M) of a solution, we need to know the moles of solute and the volume of the solution. In this case, we have a 60.0% (m/m) solution of sucrose (C12H22O11) in water with a given density at a certain temperature. Let's go step by step to find the molarity of this solution.
Step 1: Calculate the mass of the solution
To find the mass of the solution, we need to know the density of the solution and the volume. The density is given as 1.29 g/mL. Let's assume we have 100 mL of the solution since the volume is not specified in the question. Therefore, the mass of the solution can be calculated as:
Mass of solution = Volume × Density
Mass of solution = 100 mL × 1.29 g/mL
Mass of solution = 129 g
Step 2: Calculate the mass of sucrose (solute)
Since we have a 60.0% (m/m) solution, it means that 60.0 g of sucrose is present in 100 g of the solution. Using this information, we can calculate the mass of the sucrose (solute) in the solution:
Mass of sucrose = Mass of solution × Percentage of sucrose (m/m)
Mass of sucrose = 129 g × 0.60
Mass of sucrose = 77.4 g
Step 3: Calculate the moles of sucrose (solute)
To calculate the moles of solute, we need to know the molar mass of sucrose. The molar mass of C12H22O11 can be calculated as:
Molar mass of C12H22O11 = (12 × 12.01 g/mol) + (22 × 1.01 g/mol) + (11 × 16.00 g/mol)
Molar mass of C12H22O11 = 342.34 g/mol
Now, we can calculate the moles of sucrose:
Moles of sucrose = Mass of sucrose / Molar mass of C12H22O11
Moles of sucrose = 77.4 g / 342.34 g/mol
Moles of sucrose ≈ 0.23 mol
Step 4: Calculate the volume of the solution
Since we assumed the volume to be 100 mL in Step 1, we can use the same value here.
Step 5: Calculate the molarity of the solution
Finally, we can calculate the molarity using the moles of solute and the volume of the solution:
Molarity (M) = Moles of solute / Volume of solution (in liters)
Molarity = 0.23 mol / 0.1 L
Molarity = 2.3 M
Therefore, the molarity of the 60.0% (m/m) solution of sucrose in water is approximately 2.3 M.