A 0.386g sample of a protein was dissolved in water and made up to a volume of 60mL at 21.0°C. At this temperature the osmotic pressure of this solution was 1.24mmHg. Calculate the molecular mass of the protein
12682 g mol-1
95025 g mol-1
9627613 g mol-1
5706 g mol-1
125.17 g mol-1
I just want to confirm my working
1.24mmHg/760torr= 0.00163atm
21+273= 294K
0.00163= C*0.08206*294
C= 0.0000676M
moles= 0.0000676*0.06L= 0.00000405
mass= 0.386g/0.00000405mol
= 9.5*10^4g/mol
1 answer
That looks good to me. Good work.
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