A solution is prepared by dissolving 17.1grams of sucrose, c12h22o11, in 275grams of h20.
what is the molality of that solution?
I know what formula to use, but can someone please show me all the steps to get the answer. Thanks:)
17.1 g sucrose * 1 mol sucrose/ molar mass (342.34 g) = ~0.05 mol sucrose
m = mol of solute/ kg of solvent (water)
m = 0.05 mol sucrose/ 0.275 kg water = 0.18 m
Answer: 0.18 molality
-- Hope this helps future chem students :)
Ah, the sweet world of chemistry! Don't worry, I've got you covered with some sugar-coated steps to answer your question:
Step 1: Calculate the moles of sucrose (C12H22O11)
To determine the number of moles, you'll need to divide the mass of sucrose by its molar mass. The molar mass of sucrose is 342.3 g/mol.
Moles of sucrose = mass of sucrose / molar mass of sucrose
Moles of sucrose = 17.1 g / 342.3 g/mol
Step 2: Calculate the moles of water (H2O)
To calculate the moles of water, divide the mass of water by its molar mass. The molar mass of water is 18.0 g/mol.
Moles of water = mass of water / molar mass of water
Moles of water = 275 g / 18.0 g/mol
Step 3: Calculate the molality of the solution
Molality is defined as the number of moles of solute (sucrose) divided by the mass of the solvent (water) in kilograms. Since 1 kg = 1000 g, divide the moles of sucrose by the mass of water in kg.
Molality = moles of sucrose / (mass of water in g / 1000)
Molality = (17.1 g / 342.3 g/mol) / (275 g / 1000)
Step 4: Calculate the result
Now, it's time to put it all together and calculate the molality of the solution.
Molality = (17.1 g / 342.3 g/mol) / (275 g / 1000)
Molality ≈ 0.005 M
So, the molality of the solution is approximately 0.005 M. Hope you found these steps both informative and sweet!
To calculate the molality of a solution, you need to use the formula:
Molality (m) = (moles of solute) / (mass of solvent in kg)
Let's break down the steps to solve this problem:
Step 1: Determine the moles of sucrose (C12H22O11):
The molar mass of sucrose (C12H22O11) is calculated as follows:
12(Molar mass of carbon) + 22(Molar mass of hydrogen) + 11(Molar mass of oxygen)
= 12(12.01) + 22(1.01) + 11(16)
= 144.12 + 22.22 + 176
= 342.34 g/mol
Now, let's convert the grams of sucrose to moles:
Moles of sucrose = Mass of sucrose / Molar mass of sucrose
= 17.1 g / 342.34 g/mol
≈ 0.050 moles of sucrose
Step 2: Convert the mass of water to kilograms:
The mass of water is given as 275 grams. To convert it into kilograms, divide by 1000:
Mass of water in kg = 275 g / 1000
= 0.275 kg
Step 3: Calculate the molality:
Molality = Moles of solute / Mass of solvent in kg
Molality = 0.050 moles / 0.275 kg
≈ 0.182 mol/kg
Therefore, the molality of the solution is approximately 0.182 mol/kg.
If you know the formula to use you should have no trouble getting the answers.
molality = mols/kg solvent.
mols sucrose = g/molar mass
kg solvent = 0.275 kg.