A 3.50 g sugar cube (sucrose: C12H22O11) is dissolved in a 350 mL teapot containing 80∘ C water (density of water at 80∘ C = 0.975 g/mL). What is the molality of the sugar solution?

Answer: 0.04893mole/kg

Explanation:

Molality is defined as the mole of solute per kilogram of the solvent.

Sugar is the solute while water is the solvent.

Molar mass of sugar =

1*12+1*22+11*16

12+22+176=210g/mol.

Mole of solute = reacting mass / molar mass

Reacting mass 3.50g and molar mass 210g/mol

Mole = 3.50/210=0.0167moles.

For solvent

Density=mass/volume

Mass = density * volume

0.975*350

341.25g

Converting gram back to kilogram

341.25/1000 = 0.34125kg

Molality = 0.0167/0.34125

=0.04893mole/kg

Molality is 0.03 m

Explanation:

Molality indicates the moles of solute that are contained in 1000g of solvent.

We have solvent volume and density, so let's find out the mass:

Solvent density = Solvent mass / Solvent volume

Solvent density. Solvent volume = Solvent mass → 0.975 g/mL. 350 mL

Solvent mass = 341.25g

Now we need the moles of solute → mass of solute / molar mass

3.50 g / 342 g/mol = 0.0102 moles

So finally we can make a rule of three:

341.25 g of solute contain 0.0102 moles of solute

1000 g of solute, may contain (1000. 0.0102) / 341.25 = 0.03 m