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First Order Reaction

The observed ...

Question

The observed osmotic pressure for a 0.10 M solution of $F e (N H_{4})_{2} (S O_{4})_{2}$ at $25^{o} C$ is 10.8 atm. The expected and experimental (observed) values of Van't Hoff factor (i) will be respectively:

$[R = 0.082 L a t m K^{- 1} m o l^{- 1}]$

5

and

4.42

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4

and

4.00

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5

and

3.42

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3

and

5.42

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Solution

The correct option is B $5$ and $4.42$
$F e (N H_{4})_{2} (S O_{4})_{2} ⟶ 2 N H_{4}^{+} + F e^{2 +} + 2 {S O_{4}}^{2 -}$

Expected value of $i = 5$

Now, observed van't hoff factor $= \frac{P}{M R T}$

$= \frac{10.8}{0.10 \times 0.082 \times 298}$

$= 4.42$
Hence, option A is correct.

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Similar questions

0.10

F e (N H_{4})_{2} (S O_{4})_{2}

25^{o} C

[R = 0.082 {L atm K}^{- 1} {mol}^{- 1}]

0.10 M

F e (N H_{4}) 2 (S O_{4})_{2}

25 ° C

10.8 atm

[R = 0.082 L {atm K}^{- 1} {mol}^{- 1}]

5 %

25^{o} C

= 180

R = 0.0821

K^{- 1} m o l^{- 1}

The osmotic pressure of a 0.010 M MgSO₄ solution at 500 K is 0.82 atm. Calculate , the van't Hoff factor, for this MgSO₄ solution.

4.0 g

(m o l a r m a s s 246)

27^{o} C

(R = 0.082 L a t m k^{- 1} m o l^{- 1})

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