18 gram glucose (Molar mass = 180) is dissolved in 100 mol of water at 300 K. If R = 3 0.0821 L - a t m m o l - 1 K - 1 what is the osmotic pressure of solution?

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TestHub · Accepted Answer

The various quantities known to us are as follows:
$R = 0.0821 L - a t m m o l^{- 1} K^{- 1}$
$w_{2} = 18 g r a m$
Molar mass $(M_{2}) = 180$
$T = 300 K$
$V = 100 m L$
To calculate the osmotic pressure of solution, we use the following formula,
$π = \frac{w_{2} R T}{M_{2} V} = \frac{18 \times 0.0821 \times 300 \times 1000}{180 \times 100}$
$= 24.63 a t m$

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