$0.002$ molar solution of NaCl having degree of dissociation of $90 %$ at $27^{o} C$ has osmotic pressure equal to __________.

0.94

bar

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9.4

bar

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0.0934

bar

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9.4 \times 10^{- 4}

bar

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Solution

The correct option is B $0.0934$ bar
We know for dissociation,

$α = \frac{i - 1}{n - 1}$

$0.9 = \frac{i - 1}{2 - 1}$

$i = 1.9$

Now,
$π = i C R T$

$π = 1.9 \times 0.002 \times 0.082 \times 300$

$π = 0.0934 b a r$

Option C is correct.

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0.002 molar solution of NaCl having degree of dissociation of 90% at 27oC has osmotic pressure equal to __________.

The correct option is B 0.0934 barWe know for dissociation,α=i−1n−10.9=i−12−1 i=1.9Now,π=iCRTπ=1.9×0.002×0.082×300π=0.0934 barOption C is correct.

$0.002$ molar solution of NaCl having degree of dissociation of $90 %$ at $27^{o} C$ has osmotic pressure equal to __________.

The correct option is B $0.0934$ bar
We know for dissociation,

$α = \frac{i - 1}{n - 1}$

$0.9 = \frac{i - 1}{2 - 1}$

$i = 1.9$

Now,
$π = i C R T$

$π = 1.9 \times 0.002 \times 0.082 \times 300$

$π = 0.0934 b a r$

Option C is correct.