Question

A solution prepared by dissolving 8.95mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°c. Assuming the gene fragment is a non-electrolyte, determine it's molar mass.
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Answer 1

Answer:

Hi...

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.

hope it helps...❤..

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Answer 2

Answer:

good morning... how are you