Question

$18$ gram glucose (Molar mass $= 180$) is dissolved in $100 \,ml$ of water at $300\, K$. If $R = 3$ $0.0821\, L-atm\, mol^{-1}\, K^{-1}$ what is the osmotic pressure of solution ?

• A. $2.463\, atm$
• B. $24.63\, atm$
• C. $8.21\, atm$
• D. $0.821\, atm$

Answer 1

Answer: (B)

The various quantities known to us are as follows:

$R =0.0821 L$ - atm mol $^{-1} K ^{-1}$

$w _{2} =18$ gram

Molar mass $\left(M_{2}\right) =180$

$T =300 K$

$V =100 mL$

To calculate the osmotic pressure of solution, we use the following formula,

$\pi =\frac{w_{2} R T}{M_{2} V}=\frac{18 \times 0.0821 \times 300 \times 1000}{180 \times 100}$

$=24.63$ atm