Question

The nonvolatile, nonelectrolyte glucose, C6H12O6 (180.2 g/mol), is soluble in water H2O. Calculate the osmotic pressure (in atm) generated when 11.9 grams of glucose are dissolved in 217 mL of a water solution at 298 K.

Answer 1

Answer: The osmotic pressure of the solution is 7.44 atm

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = given mass of glucose = 11.9 g

$M_{solute}$ = molar mass of glucose = 180.2 g/mol

$V_{solution}$ = Volume of solution = 217 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$\pi=1\times \frac{11.9\times 1000}{180.2\times 217}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=7.44atm$

Hence, the osmotic pressure of the solution is 7.44 atm