Question

a solution containing 10g/l sucrose has an osmotic pressure of 0.66 atm at 273 k. calculate the value of constant R

Answer 1

10g sucrose contains 1L of solution .
weight of sucrose in 1L = 10g
molecular weight of sucrose ,(C₁₂H₂₂O₁₁) = 342 g/mol
∴ mole of sucrose = 10/342 = 0.02923
so, concentration of sucrose in solution = 0.02923mol/1L = 0.02923 M

we know, Osmotic pressure , π = CRT
here C is concentration, R is universal gas constant and T is temperature in Kelvin.
Given, C = 0.02923M , π = 0.66 atm and T = 273K
Now, R = π/CT = 0.66/0.02923 × 273 = 0.08270 atm.L/mol/K

Hence, R = 0.08270 atm.L/mol/K

Answer 2

Answer :

• R = $\bold{0.0827\:L\:atm\:K^{-1}\:mol^{-1}}$

Step-by-step explanation :

Given that,

• Concentration (C) = 10 g/litre = 10/342 moles/litre (Molar mass of sucrose = 342 g/mol).

• Osmotic pressure ($\pi$) = 0.66 atm.

• Temperature (T) = 273 K.

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Using the equation, $\bold{\pi=CRT}$, we get,

$\implies\bold{R=\dfrac{\pi}{CT}}$

Applying the given values, we get,

$\implies\bold{R=\dfrac{0.66\:atm}{\frac{10}{342}mol\:L^{-1}\times{}273\:K}}$

$\implies\bold{R=\dfrac{0.66\times{}342}{10\times{}273}L\:atm\:K^{-1}\:mol^{-1}}$

$\implies\bold{R=0.0827\:L\:atm\:K^{-1}\:mol^{-1}.}$