Calculate the molar mass of nonelectrolyte solute when 6 gram of it is dissolved in 1 dm 3 water has osmotic pressure 2.4 atm at 300 K ( R = 0.0821 atm dm 3 K − 1 mol − 1 )

Question

TestHub · Accepted Answer

The osmotic pressure is given by the formula:

$π = CRT = \frac{W RT}{M V}$

Rearranging the formula to solve for the molar mass $M$ :

$M = \frac{W RT}{πV}$

Substituting the given values $W = 6 g$ , $R = 0.0821 atm dm^{3} K^{- 1} mol^{- 1}$ , $T = 300 K$ , $π = 2.4 atm$ , and $V = 1 dm^{3}$ :

$M = \frac{6 \times 0.0821 \times 300}{2.4 \times 1}$

$M = \frac{147.78}{2.4}$

$M = 61.575 g mol^{- 1}$

Rounding to two decimal places, the molar mass is $61.58 g mol^{- 1}$ .

Answer: $61.58 g mol^{- 1}$

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