Question

at 300 Kelvin 30 gram of C6H12O6 present in a litre of its solution has an osmotic pressure of 4.98 bar if the osmotic pressure of a glucose solution is 1.52 bar at the same temperature what would be its concentration​

Answer 1

Here we have given

π1= 4.98

π2 = 1.52

C1 = 30/180

C2 = ? (we have to find)

Now according to van’t hoff equation

Π = CRT

Putting the values in above equation,we get

4.98 = 30/180RT ------------------------1

1.52 = c2RT       ------------------------2

Now dividing equation 2 by 1 ,we get

(c2 x 180) / 30 = 1.52 / 4.98

or

c2 = 0.050833..

Therefore concentration of 2nd solution is  0.050833M

but the question in my textbook has different values... it has 36 grams instead of 30.. Here's the answer for that

π1= 4.98

π2 = 1.52

C1 = 36/180

C2 = ? (we have to find)

Now according to van’t hoff equation

Π = CRT

Putting the values in above equation,we get

4.98 = 36/180RT ------------------------1

1.52 = c2RT       ------------------------2

Now dividing equation 2 by 1 ,we get

(c2 x 180) / 36 = 1.52 / 4.98

or

c2 = 0.0061

Therefore concentration of 2nd solution is 0.0061 M