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# Answer to Question #194143 in Chemistry for LAGUNA STATE POLYT

Question #194143

A solution contains 68.4 g of sucrose (C12H22O11) dissolved in 250 g of water. What

is the molality of the solution?

1
2021-05-17T04:08:49-0400

Solution:

Molality of a solution is the moles of solute divided by the kilograms of solvent.

Molality = Moles of solute (mol) / Mass of solvent (kg)

solute = sucrose (C12H22O11)

solvent = water (H2O)

Mass of H2O = 250 g = 0.25 kg

Mass of C12H22O11 = 68.4 g

The molar mass of C12H22O11 is 342.3 g/mol

Therefore,

(68.4 g C12H22O11) × (1 mol C12H22O11 / 342.3 g C12H22O11) = 0.1998 mol = 0.2 mol C12H22O11

Moles of C12H22O11 = 0.2 mol

Thus,

Molality of the solution = (0.2 mol) / (0.25 kg) = 0.8 mol/kg = 0.8 m

Molality of the solution = 0.8 m

Answer: The molality of the solution is 0.8 m

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