Question

Calculation osmotic pressure of the solution of 6.5 g of glucose (molar mass=180 g mol-1) dissolved in 200 ml of water at 300k

Answer 1

See attachment it may help

Answer 2

The osmotic pressure of glucose solution is 4.45 atm

Explanation:

To calculate the osmotic pressure, we use the equation:

$\pi=iMRT$

or,  

$\pi=i\times \frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}}\times RT$

where,

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

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

Given mass of glucose = 6.5 g

Molar mass of glucose = 180 g/mol

Volume of solution = 200 mL

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

T = temperature = 300 K

Putting values in above equation, we get:

$\pi=1\times \frac{6.5\times 1000}{180\times 200}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 300K\\\\\pi=4.45atm$

Learn more about osmotic pressure:

https://brainly.in/question/6671539

https://brainly.in/question/3245367

#learnwithbrainly