At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height ( h ) of the solution (density = 1.00 g cm − 3 ) where h is equal to 2.00 cm . I

Question

At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height ( h ) of the solution (density $= 1.00 g cm^{- 3}$ ) where h is equal to 2.00 cm . If the concentration of the dilute solution of the macromolecule is $2.00 g dm^{- 3}$ , the molar mass of the macromolecule is calculated to be $X \times 1 0^{4} g mol^{- 1}$ . The value of $X$ is . Use : Universal gas constant $(R) = 8.3 J K^{- 1} mol^{- 1}$ and acceleration due to gravity $(g) = 10 m s^{- 2}$

Answer:

2.49

Solution:

$π = ρ g h = 1 0^{3} \times 10 \times 2 \times 1 0^{- 2} Pascal = 200 Pascal π = CRT 200 = \frac{2}{M} \times 1000 \times 8.3 \times 300 M = 24900 = 2.49 \times 1 0^{4} g / mol X = 2.49$

TestHub · Accepted Answer

​ π = ρ g h = 1 0 3 × 10 × 2 × 1 0 − 2 Pascal = 200 Pascal π = CRT 200 = M 2 ​ × 1000 × 8.3 × 300 M = 24900 = 2.49 × 1 0 4 g / mol X = 2.49 ​

Chemistry - Liquid Solution Question with Solution | TestHub

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