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Calculate the molaity of a solution that is prepared by dissolving 50.4g sucrose (C12H22O11) in 0.332kg of water

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Answer:

0.443 mol/kg.

Step-by-step explanation:

The formula for molal concentration (b) is


b = \frac{\text{moles of solute}}{\text{kilograms of solvent}}


\text{Moles of solute} = \text{50.4 g sucrose} * \frac{ \text{1 mol sucrose}}{\text{342.30 g sucrose} } = \text{0.1472 mol sucrose}


b = \frac{ \text{0.1472 mol}}{ \text{0.332 kg}} = \text{0.443 mol}\cdot\text{kg}^(-1)

User Syed Ayesha Bebe
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