Two solutions of glucose have osmotic pressures $1.5$ and $2.5 a t m .$ $1 L$ of first is mixed with $2 L$ of second solution, the osmotic pressure of the resultant solution is:

2.5 + 1.5 a t m

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2.5 - 1.5 a t m

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\frac{2.5 + 1.5}{2} a t m

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2.16 a t m

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Solution

The correct option is D $2.16 a t m$
$π_{1} = C_{1} R T$
$1.5 = \frac{n_{1}}{1} \times R T$
$∴ n_{1} = \frac{1.5}{R T}$
$π_{2} = C_{2} R T$
$2.5 = \frac{n_{2}}{2} \times R T$
$∴ n_{2} = \frac{5}{R T}$
Now, $π_{m i x} = C_{m i x} R T$
$= (\frac{n_{1} + n_{2}}{v_{1} + v_{2}}) R T$
$= \frac{\frac{1.5}{R T} + \frac{5}{R T}}{3} \times R T$
$= \frac{6.5}{3 R T} \times R T$
$π_{m i x} = 2.16 a t m =$ osmostic pressure of mixture

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Two solutions of glucose have osmotic pressures $1.5$ and $2.5 a t m .$ $1 L$ of first is mixed with $2 L$ of second solution, the osmotic pressure of the resultant solution is: