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

Derivation of Kp and Kc

An aqueous so...

Question

An aqueous solution of urea has freezing point of $- {0.52}^{o} C$ . Assume molarity and molality by same. $K_{f}$ for $H_{2} O$ is 1.86 K $m o l^{- 1} k g$ . The osmotic pressure of solution at $27^{o} C$ in atm is (nearest integer)

Open in App

Solution

$∵ Δ T = \frac{1000 \times K_{f} \times w}{m \times W}$
$= K_{f} \times m o l a l i t y$
$∴ M o l a l i t y = \frac{0.52}{1.86} = m o l a l i t y (\frac{n}{V})$ (Given)
Now, $π V = n S T$
$∴ π = (\frac{n}{V}) S T = (0.52 / 1.86) \times 0.0821 \times 300$
$= 6.886 \approx 7 a t m$

Suggest Corrections

Similar questions

25^{o} C

K_{f}

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

2.0

300

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

R = 0.08251 L a t m K^{- 1} m o l^{- 1}

- {0.52}^{o} C .

K_{f} = {1.86}^{0} C / m

0.1 M H A (a q)

- {0.2046}^{\circ} C

p H

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

N a H S O_{4}

- 0.372^{o} C

H S O_{4}^{-} \to H^{+} + S O_{4}^{- 2}

N a H S O_{4}

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

Join BYJU'S Learning Program