Question

A solution prepared by dissolving 8.95 mg of gene fragment in 35 ml of water has an osmotic pressure of0.335 torr at 25 C .Assuming that the gene fragment is a non electrolyte ,calculate its molar mass.

Answer 1

Dear student!

We know that the molar concentration of the gene fragment is given as:

8.95 mg in 35 ml water 

Or, 8.95 x 10-3 gm in 35/ 1000 L of water

If M is the molar mass of the gene fragment then, 

Molar concentration ,C=  (8.95 x 10-3 gm/ M) x(1000/35)  gMol/L

and so, molar concentration =  0.255 /M  gMol/L

Now, ∏ = CRT

Here, ∏ =  0.335 torr = 0.335/760 atm  

R =0.082 L atm/Kmol 

T = 250C = 273 +25 =298 K

So, putting the values we get,

0.335/760 = (0.255 /M  ) x 0.082 L  x 298 

Or. M = 14176.02

Answer 2

Answer: 15641g/mol

Explanation: $\pi =CRT$

$\pi$ = osmotic pressure  = 0.335 torr=0.0004atm

C= concentration  in molarity.

R= solution constant  = 0.0821Latm/molK

T= temperature  = 25C= (25+273)=298K

8.95mg=0.00895g of gene is dissolved in 35 ml of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$0.0004atm=\frac{0.00895\times 1000}{M}\times {35}0.0821\times 298K$

$M= 15641g/mol$