Calculate the molality of each of the following solutions: (a) 14.3 g of sucrose (C12H22O11) in 685 g of water, (b) 7.15 moles

Question

Calculate the molality of each of the following solutions:
(a) 14.3 g of sucrose (C12H22O11) in 685 g of water,
(b) 7.15 moles of ethylene glycol (C2H6O2) in 3505 g of water.

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Nem 3 years 2021-07-17T14:08:54+00:00 1 Answers 522 views 0

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  1. minhkhoi
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    2021-07-17T14:09:58+00:00
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    Answer:

    (a) The molality of this solution is 0.0613\frac{moles}{kg}

    (b)The molality of this solution is 2.04\frac{moles}{kg}

    Explanation:

    The molality (m) of a solution is defined as the number of moles of solute present per kg of solvent.

    The Molality of a solution is determined by the expression:

    Molality=\frac{number of moles of solute}{kilograms of solvent}

    Molality is expressed in units \frac{moles}{kg}

    (a) You have 14.3 g of sucrose (C₁₂H₂₂O₁₁), the solute. With the molar mass of sucrose being 342 \frac{g}{mole}, then 14.3 grams of the compound represents the following number of moles:

    14.3 grams*\frac{1 mole}{342 grams} = 0.042 moles

    Having 685 g= 0.685 kg (being 1000 g= 1 kg) of water, the solvent, molality can be calculated as:

    molality=\frac{0.042 moles}{0.685 kg}

    Solving:

    molality= 0.0613\frac{moles}{kg}

    The molality of this solution is 0.0613\frac{moles}{kg}

    (b) In this case you have 7.15 moles of ethylene glycol (C₂H₆O₂), the solute, in 3505 g (equal to 3.505 kg) of water, the solvent, molality can be calculated as:

    molality=\frac{7.15 moles}{3.505 kg}

    Solving:

    molality= 2.04\frac{moles}{kg}

    The molality of this solution is 2.04\frac{moles}{kg}

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