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

Step-by-step explanation:

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

`\pi=iMRT`

or,

`\pi=i* \frac{m_(solute)* 1000}{M_(solute)* V_(solution)\text{ (in mL)}}}* 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* (11.9* 1000)/(180.2* 217)* 0.0821\text{ L. atm }mol^(-1)K^(-1)* 298K\\\\\pi=7.44atm`

Hence, the osmotic pressure of the solution is 7.44 atm