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19.5 g of CH(2)FCOOH is dissolved in 500...

19.5 g of `CH_(2)FCOOH` is dissolved in 500 g of water. The depression in the freezing point observed is `1.0^(@)C`. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid. `K_(f)` for water is `"1.86 K kg mol"^(-1)`.

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To solve the problem, we will go through the following steps: ### Step 1: Calculate the molar mass of fluoroacetic acid (CH₂FCOOH) The molecular formula of fluoroacetic acid is CH₂FCOOH. We can calculate its molar mass by adding the atomic masses of its constituent atoms: - Carbon (C): 12.01 g/mol (2 atoms) - Hydrogen (H): 1.008 g/mol (2 atoms) - Fluorine (F): 19.00 g/mol (1 atom) - Oxygen (O): 16.00 g/mol (2 atoms) ...
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9.5 g of CH_2 FCOOH is dissolved in 500 g of water. The depression in the freezing point observed is 1.0^@C . Calculate the van't Hoff factor and dissociation constant of fluroacetic acid.

19.5g of CHM_(2)FCOOH is dissolved in 500g of water . The depression in the freezing point of water observed is 1.0^(@)C . Calculate the Van't Hoff factor and dissociation constant of fluoroacetic acid.

Knowledge Check

  • The freezing point of 0.2 molal K_(2)SO_(4) is -1.1^(@)C . Calculate van't Hoff factor and percentage degree of dissociation of K_(2)SO_(4).K_(f) for water is 1.86^(@)

    A
    `97.5`
    B
    `90.75`
    C
    `105.5`
    D
    `85.75`
  • A 0.5% aqueous solution of KCl was found to freeze at -0.24^(@)C . Calculate the Van,t Hoff factor and degree of dissociation of the solute at this concentration. ( K_(f) for water = 1.86 K kg mol ^(-1) )

    A
    1.92 and 19.2%
    B
    0.92 and 92%
    C
    1.92 and 92%
    D
    9.2 and 9.2%
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    0.6 mL of acetic acid (CH_(3)COOH) having density 1.06 g mL^(-1) is dissolved in 1 L of water. The depression in freezing point observed for this strength of acid was 0.0205^(@)C .Calculate the Van't Hoff factor and dissociation constant of the acid. K_(f) for H_(2)O=1.86 K kg ^(-1) "mol"^(-1))

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