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Two grams of benzoic acid (C(6)H(5)COOH)...

Two grams of benzoic acid `(C_(6)H_(5)COOH)` dissolved in `25.0 g` of benzene shows a depression in freezing point equal to `1.62 K`. Molal depression constant for benzene is `4.9 K kg^(-1)"mol^-1`. What is the percentage association of acid if it forms dimer in solution?

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To solve this problem, we need to determine the percentage association of benzoic acid when it forms a dimer in benzene. Here are the steps to solve the problem: ### Step 1: Determine the molality of the solution First, we need to calculate the molality of the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent. Given: - Mass of benzoic acid (solute), \( W_2 = 2 \, \text{g} \) - Mass of benzene (solvent), \( W_1 = 25 \, \text{g} = 0.025 \, \text{kg} \) ...
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2 g of benzoic acid (C_6H_5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg "mol"^(-1) . What is the percentage association of acid if it forms dimer in solution ?

2 g of benzoic acid (C_(6) H_(5) COOH) dissolved in 25 g of benzene shows a depressionc in freezing point equal to 1.62 K. Molal depression constant for benzene is "4.9 K kg mol"^(-1) . What is the percentage association of acid if it forms dinner in solution? How many mL of "0.1 M HCl" are required to react completely with 1 g mixture of Na_(2) CO_(3) and NaHCO_(3) containing equimolar amounts of both?

Knowledge Check

  • 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg moll. The molar mass of the solute is

    A
    256 kg/mol
    B
    256 g mol
    C
    256 g/mol
    D
    256 mg/mol

  • 1.00 g of q non - electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 k. The freezing point depression constant of benzene is 5.12 k kg mol^(-1) The molar mass of the solute is

    A
    256 kg/mol
    B
    256 g mol
    C
    256 g / mol
    D
    256 mg /mol

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    2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant (K_(f)) of benzene is "4.9 K kg mol"^(-1) . What is the percentage association of the acid if it forms dimer in the solution?

    2g of benzoic acid dissolved in 25g of C_(6)H_(6) . Shows a depression in freezing point equal to 1.62K . Molal depression constant of C_(6)H_(6) is 4.9K kg "mol"^(-1) . What is the percentage association of acid if it forms double molecules in solution?

    (i) 1.51 g of NaCl was dissolved in 500g of water and the elveation in boiling point observed to be 0.0514^(@)C . Calculate the van't . Hoff factor K_(b) for water 0.514 K kg/mol. (ii) 1.1 g of CopCl_(2) 6NH_(3) (mot wt. =267) was dissolved at 100 g"of" H_(2)O . The freezing point of solution is 0.29^(@)C . How many moles of solute particies exist in solution for eac mol of soluta introduced ? K_(f) for H_(2)O=1.86^(2) Cm^(-1) (iii) 2g of bezoic acid dissolved in 25g of C_(6)H_(5)^(-) . shows a deptession in freezing pont equal to 1.62K . Molal depression constant of benzene is 4.9 K mol/kg . What is the percentage association of acid if it dimenzes in solution.

    Molecules of benzoic acid (C_(6)H_(5)COOH) dimerise in benzene. ‘w’ g of the acid dissolved in 30g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimer in the solution is 80, then w is : (Given that K_(f)=5K Kg "mol"^(-1) Molar mass of benzoic acid =122 g "mol"^(-1) )

    1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is "5.12 K kg mol"^(-1) . Find the molar mass of solute.

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