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A 0.025 m sol...

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A 0.025 m solution of monobasic acid had a freezing point of -0.060^° C. The pKa for the acid is(K_b=1.86 K kg/mol)- 1) 1.2, 2) 2, 3) 2.5, 4) 5.7

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0.025 M

aqueous solution of a monobasic acid has a freezing point of

- {0.06}^{\circ} C

.
Calculate acid dissociation constant

(K_{a})

(K_{f})_{H_{2} O} = 1.86 K k g m o l^{- 1}

Assume molality = molarity for a dilute solution

Q. A 0.025 M solution of monobasic acid has a freezing point of

- {0.60}^{o} C

K_{f}

H_{2} O =

1.86 K m o l a l i t y^{- 1}

. Assume molality equal to molarity.

K_{a}

for the acid is:

Q. The freezing point of a

0.025 M

solution of a weak acid

H A

- {0.060}^{\circ} C

. The percent dissociation is:

K_{b} (H_{2} O) = 1.86 K K g {m o l}^{- 1}

0.1

molal solution of a monobasic acid is

45 %

ionized. Calculate the depression in freezing point:

(Molecular weight of the acid

= 300

K_{f} = 1.86 K {m o l}^{- 1} k g

Q. 1g of a monobasic acid in 100 g water lowers the freezing point by

{0.168}^{o}

C. If 0.2 g of the same acid requires 15 ml of

\frac{N}{10}

alkali solution for complete neutralization, the degree of dissociation of acid is nearly:

K_{f}

H_{2} O = 1.86 K k g m o l^{- 1})

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