Question

300 g sucrose was dissolved in $1d{m}^3$ solution. If the osmotic pressure is $2.19\times {10}^{x}N{m}^{-2}$ at ${27}^{o}C$. Find the value of x.

• A. $x=4$
• B. $x=5$
• C. $x=6$
• D. $x=7$

Answer 1

The correct option is C $x=6$

Osmosis is the diffusion of a fluid through a semipermeable membrane. When a semipermeable membrane (animal bladders, skins of fruits and vegetables) separates a solution from a solvent, then only solvent molecules are able to pass through the membrane. The osmotic pressure of a solution is the pressure difference needed to stop the flow of solvent across a semipermeable membrane. The osmotic pressure of a solution is proportional to the molar concentration of the solute particles in solution.

$\pi =\frac{nRT}{V}=MRT$

Here, $\pi$ is osmotic pressure.

$R=8.314$ is the gas constant.

$T$ is temperature.

$M$ is molarity of solute and $n$ is number of moles of solute present.

Number of moles of sucrose will be $n=\frac{300}{342}=0.877$

Volume $V=1d{m}^3=0.001m^3$

Temperature will be $T=27+273=300$

Osmotic pressure will be $\frac{0.877\times 8.314\times 300}{0.001}=2187413.4=2.18\times {10}^6$

$⟹x=6$