You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Depression in Freezing Point

An aqueous so...

Question

An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7g of water freezes at 272.813 K. If the value of $K_{f}$ for water is 1.86 K/m, what is the molecular mass of the monobasic acid.

Open in App

Solution

As we know $T_{f} = i K_{f} m$ $m =$ molality; $K_{f} = 1.86$
$\Rightarrow 272.81 = 2 \times 1.86 \times m$
$\Rightarrow m = 73.33$
$\Rightarrow 73.33 = \frac{0.1}{M} \times \frac{21.7}{1000}$
$\Rightarrow H = 0.6628 g m$ .

Suggest Corrections

Similar questions

An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7g of water freezes at 272.813 K. If the value of K_f for water is 1.86 K/m, what is the molecular mass of the monobasic acid?

Q. 1.0 g of a weak monobasic acid (

H A

), when dissolved in 150 mL water, lowers the freezing point by

{0.186}^{o} C

. Also 1.0 g of the same acid required 125 mL of a 0.10N Nate solution for complete neutralisation. Determine dissociation constant

(k_{a})

of the weak acid.

k_{f}

of water is 1.86 K kg

m o l^{- 1}

Q. A

0.1

molal solution of a monobasic acid is

45 %

ionized. Calculate the depression in freezing point:

(Molecular weight of the acid

= 300

g/mol,

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

)

Q. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If

K_{f}

for water is

1.86^{o} C / m

, the freezing point of the solution will be:

Q. A

0.025 M

aqueous solution of a monobasic acid has a freezing point of

- {0.06}^{\circ} C

.
Calculate acid dissociation constant

(K_{a})

Given:

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

Assume molality = molarity for a dilute solution

Join BYJU'S Learning Program

An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7g of water freezes at 272.813 K. If the value of Kf for water is 1.86 K/m, what is the molecular mass of the monobasic acid.

As we know Tf=iKfm m= molality; Kf=1.86⇒272.81=2×1.86×m⇒m=73.33⇒73.33=0.1M×21.71000⇒H=0.6628gm .

An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7g of water freezes at 272.813 K. If the value of $K_{f}$ for water is 1.86 K/m, what is the molecular mass of the monobasic acid.

As we know $T_{f} = i K_{f} m$ $m =$ molality; $K_{f} = 1.86$
$\Rightarrow 272.81 = 2 \times 1.86 \times m$
$\Rightarrow m = 73.33$
$\Rightarrow 73.33 = \frac{0.1}{M} \times \frac{21.7}{1000}$
$\Rightarrow H = 0.6628 g m$ .