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Answer :
Answer:
to calculate the molarity of the said sucrose,
firstly calculate the moles
which is = Molecular weight of C12H22O11 = 342g/mol
then
moles = 139/342
= 0.41 moles
to calculate Molarity now
Molarity= moles of the solute/volume of solution in liter
=0.41/2.60
=0.158M
Explanation:
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Final answer:
The molality of a 3.1416 M aqueous solution of sucrose with a density of 1.5986 is calculated to be approximately 6.005 mol/kg, using steps that include finding the total mass of the solution, the mass of sucrose, and then the mass of water to determine the molality.
Explanation:
To calculate the molality of a 3.1416 M aqueous solution of sucrose with a density of 1.5986, we first convert the density to grams per liter and then find the mass of the solute and the solvent. The molar mass of sucrose, C₁₂H₂₂O₁₁, is 342.297 g/mol. To begin, we would calculate the total mass of the solution per liter:
1.5986 g/mL × 1000 mL/L = 1598.6 g/L solution
Then, knowing the molarity, we can find the mass of sucrose per liter of solution:
3.1416 mol/L × 342.297 g/mol = 1075.4 g sucrose/L solution
We subtract the mass of sucrose from the total mass of the solution to find the mass of water:
1598.6 g solution - 1075.4 g sucrose = 523.2 g water
Convert this mass to kilograms (since molality requires kilograms of solvent):
523.2 g water × 1 kg/1000 g = 0.5232 kg water
Finally, we calculate the molality (moles of solute per kilogram of solvent):
3.1416 mol sucrose / 0.5232 kg water = 6.005 mol/kg
