Suppose 3.005 g of a nonvolatile solute is added to 20.02 g of water (the solvent), and the boiling point increases from 100.000 OC to 101.2

Question

Suppose 3.005 g of a nonvolatile solute is added to 20.02 g of water (the solvent), and the boiling point increases from 100.000 OC to 101.286 OC. Determine the TB, molality, moles, and molecular weight for the solute if kb for water is 0.512 OC/m. Report each value using the correct number of significant digits. Refer to Example 1.2 and pages 3-4 in the chapter 1 notes for general chemistry 1 to understand significant figures.

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Aubrey 7 mins 2022-06-23T10:53:17+00:00 1 Answer 0 views 0

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    2022-06-23T10:54:26+00:00

    Answer:

    * [tex]\Delta T_B=1.286\°C[/tex]

    * [tex]m=2.5m[/tex]

    * [tex]n=0.05mol[/tex]

    * [tex]M=59.76g/mol[/tex]

    Explanation:

    Hello,

    In this case, considering the boiling point rise problem, we consider its appropriate equation:

    [tex]\Delta T_B=imK_b[/tex]

    Whereas i is the van’t Hoff factor that for this nonvolatile solute is 1, m is the molality, Kb the boiling point constant of water as it is the solvent and ΔT the temperature difference. In such a way, with the given information we obtain:

    – ΔT:

    [tex]\Delta T_B=101.286\°C-100.000\°C\\\\\Delta T_B=1.286\°C[/tex]

    – Molality (mol/kg):

    [tex]m=\frac{\Delta T_B}{i*K_b}=\frac{1.286\°C}{1*0.512\°C/m}\\ \\m=2.5m[/tex]

    – Moles for 20.02 g (0.02002 kg) of water:

    [tex]n=2.5mol/kg*0.02002kg\\\\n=0.05mol[/tex]

    – Molar mass:

    [tex]M=\frac{mass}{moles}=\frac{3.005g}{0.050mol} \\\\M=59.76g/mol[/tex]

    Best regards.

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