Osmotic pressure of $40 %$ $(w / V)$ urea solution is $1.64$ $a t m$ . and that of $3.42 %$ $(w / V)$ cane sugar solution is $2.46$ $a t m$ . When equal volumes of the given two solutions are mixed, the osmotic pressure of the resulting solution will be?

1.64

a t m

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2.46 a t m

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4.10 a t m

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2.05 a t m

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Solution

The correct option is D $2.05 a t m$
When equal volumes of two solutions are mixed, its new osmotic pressure will become the average of the osmotic pressure of the solutions. It is because when volume is doubled, its concentration becomes halved.

$∴$ New osmotic pressure= $\frac{P_{1} + P_{2}}{2} = \frac{1.64 a t m + 2.46 a t m}{2} = 2.05 a t m$

Hence, the correct option is $D$

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Q. The osmotic pressure of 40% (weight/volume) urea solution is 1.64 atm and that of 3.42% (weight/volume) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is:

Q. Osmotic pressure of 40% (mass/vol.) urea solution is 1.64 atm and that of 3.42% (mass/vol.) canesugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is :

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Osmotic pressure of 40% (w/V) urea solution is 1.64 atm. and that of 3.42% (w/V) cane sugar solution is 2.46 atm. When equal volumes of the given two solutions are mixed, the osmotic pressure of the resulting solution will be?

Osmotic pressure of $40 %$ $(w / V)$ urea solution is $1.64$ $a t m$ . and that of $3.42 %$ $(w / V)$ cane sugar solution is $2.46$ $a t m$ . When equal volumes of the given two solutions are mixed, the osmotic pressure of the resulting solution will be?