Question

if 3 gram of glucose is dissolved in 60 gram of water at 15 degree Celsius then the Osmotic pressure of solution will be

Answer 1

HEY MATE!!! ❤❤❤HERE'S UR ANSWER!!! ❤❤




Let density of solution is d g/ml
Then, volume of solution = density of solution × mass of solution
= d × (mass of solvent + mass of solute )
= d × (3 + 60)
= 63d mL
so, concentration of solute = number of mole of solute/volume of solution
= {3/180} × 1000/63d = 300/{18 × 63d} mol/Litre [ molar mass of Glucose = 180g/mol]

Now, Osmotic pressure ,π = CRT
Where, C is concentration of solute in solution.
∴ π = 300/{18 × 63d} × 0.082 × (273 + 15) [ ∵ T = 15°C ]
= 300 × 0.082 × 288/(18 × 63d)
= 6.247/d atm




HOPE THIS HELPS YOU!!!!❤❤❤

Answer 2

AnsweR :-

If 3 grams of glucose is dissolved in 60 grams of water at 15 degree then osmotic pressure os

$\pi = crt$

$\frac{3 \times 1000}{180 \times 60} \times 0.0821 \times 288 \\ \\ = 6.56atm$