A solution is prepared by dissolving 17.1grams of sucrose, c12h22o11, in 275grams of h20.
what is the molality of that solution?
how would i do this???
To find the molality (m), we need to first determine the moles of sucrose (C12H22O11) and the mass of water (H2O) in the solution.
Step 1: Calculate the moles of sucrose (C12H22O11)
The molar mass of sucrose (C12H22O11) is the sum of the atomic masses of all the elements it contains. You can find this information in the periodic table or using online resources.
Molar mass of C12H22O11:
(12.01 g/mol * 12) + (1.008 g/mol * 22) + (16.00 g/mol * 11) = 342.3 g/mol
Now, we can use this molar mass to calculate the moles of sucrose:
Moles of sucrose = Mass of sucrose / Molar mass of sucrose
Moles of sucrose = 17.1 g / 342.3 g/mol
Step 2: Convert the mass of water (H2O) to kilograms
The formula for molality (m) is moles of solute / mass of solvent in kilograms. So, we need to convert the mass of water from grams (g) to kilograms (kg):
Mass of water (H2O) = 275 g
Mass of water (H2O) in kilograms = 275 g / 1000
Step 3: Calculate the molality (m)
Now that we have the moles of sucrose and the mass of water in kilograms, we can calculate the molality (m):
Molality (m) = Moles of sucrose / Mass of water (H2O) in kilograms
Substitute the values we have determined:
Molality (m) = (17.1 g / 342.3 g/mol) / (275 g / 1000 kg)
Now, perform the calculations:
Molality (m) = 0.05 mol / 0.275 kg
Molality (m) = 0.182 mol/kg
Therefore, the molality of the solution is 0.182 mol/kg.
Use the definition of molality:
molality= molessucrose/kgsolution=
=gramssucrose/(mw sucrose*.275)
figure the mw of sucrose first