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anonymous

  • one year ago

**Please Help Me**A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

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  1. anonymous
    • one year ago
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    @dan815 @Luigi0210 @zepdrix @welshfella @BTaylor @Cuanchi @Mehek14 @eyesac @aaronq

  2. anonymous
    • one year ago
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    • one year ago
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    • one year ago
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    • one year ago
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  6. aaronq
    • one year ago
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    use the formula \(\Delta T=i*m*K_f\) where \(\Delta T\) is the change in temp from the pure solvent's f.p. i is the van't hoff constant, 1 in the case for sugar m is the molality of the solution \\(K_f\) is the f.p. depression constant

  7. anonymous
    • one year ago
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    wait don't we got to find the morality first @aaronq

  8. aaronq
    • one year ago
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    yup

  9. aaronq
    • one year ago
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    then use the formula

  10. anonymous
    • one year ago
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    Can you setup how to find the moles for me

  11. anonymous
    • one year ago
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    @aaronq

  12. aaronq
    • one year ago
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    divide the mass by the molar mass of glucose (which is 180 g/mol)

  13. anonymous
    • one year ago
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    Ok I have 0.15996

  14. anonymous
    • one year ago
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    -0.15996* @aaronq

  15. aaronq
    • one year ago
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    what is that?

  16. anonymous
    • one year ago
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    Delta T @aaronq

  17. aaronq
    • one year ago
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    alright

  18. anonymous
    • one year ago
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    So that is the freezing point

  19. anonymous
    • one year ago
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    @aaronq

  20. aaronq
    • one year ago
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    idk, it mean it sounds right. you found moles, then molality, then multiplied by the constant? if so, then it should be right.

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