The osmotic pressure of a solution containing 0.10g of haemoglobin in 10 cm of solution is 30 Torr at 300 K. The molar mass of haemoglobin is nearly

Options

A. `4500g mol^(-1)`
B. `5452 g mol^(-1)`
C. `6232 g mol^(-1)`
D. `6932 g mol^(-1)`

Answer

C

Explanation

`M_(2)= ((m_(2))/(V)) ((RT)/(pi))=[(0.10g)/(10.0 xx 10^(-3)L)] [((0.082L atm K^(-1) mol^(-1))(300K))/((30//760)atm)]`
`=8.2 xx 760g mol^(-1)= 6232g mol^(-1)`