Question

At what temperature $10\%\left(w/V\right)$ aqueous solution of glucose (mol mass is 180 g/mol) will exhibit an osomotic pressure of $16.42atm$?

• A. ${360}^{\circ }C$
• B. ${180}^{\circ }C$
• C. $360K$
• D. $180K$

Answer 1

The correct option is C $360K$

$10\%\left(w/V\right)$ solution means $10g$ of glucose is present in $100mL$ of solution
Osmotic pressure:

$\pi =\frac{n}{V}\times RT\pi \times V=n\times R\times T$

Here n is number of moles

$n=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{10}{180}$

$V=100mL=0.1LR=0.0821Latmmo{l}^{-1}{K}^{-1}\pi =16.42atm$

Putting the values :

$16.42\times 0.1=\frac{1}{18}\times 0.0821\times TT=360K$