Question

In: Chemistry

To make sweet tea, a cook dissolved 147.023 grams of sugar (C6H6O2, FW = 110 g/mol)...

To make sweet tea, a cook dissolved 147.023 grams of sugar (C6H6O2, FW = 110 g/mol) in 5.07 L of water at 30.71 °C. What is the molality of this sugar solution?

Solutions

Expert Solution

Volume of water = 5.07 L = 5070 mL

density of water = 0.997 g/mL

mass of 1 mL of water = 0.997 grams

mass of 5070 mL of water = ?

                                           = 0.997x5070/1= 5054.79 grams

mass of water ( solvent ) = 5054.79 grams = 5.0548 kg

Mass of sugar =(s0lute) = 147.023 grams

Formula mass of SUgar = 110 gram/mole

number of moles of sugar = mass of sugar/molar mass = 147.023/110= 1.3366 moles

number of moles of sugar = 1.337 moles

Molality = number of moles of solute/mass of solvent in Kg

Molality = 1.337/5.0548 = 0.264M

Molality of the sugar solution = 0.264M


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