Answer:
Highest osmotic pressure that membrane may experience is
' =58.638 atm
Explanation:
Suppose sea-water taken is M= 1 kg
Density of water = 1000 kg/m3
Therefore Volume of water= Mass,M/Density of water
V= 1 kg/(1000 kg/m3)
V= 10-3 m3= 1 Litre
Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water
Mass present of NaCl= m= 0.035*1000 g
m= 35 g
Since molecular weight of NaCl= 58.44 g/mol =M.W.
Thus its Number of moles of Nacl= m/M.W
nNaCl= 35g/58.44 gmol-1
= 0.5989 mol
ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres
C= 0.5989mol/ 1L
=0.5989 M
Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2
And osmotic pressure = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)
Putting in equation 1 ,we get = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K
=29.319 atm
Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum
Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)
Thus New concentration of NaCl C'= 2*C
C'=2*0.5989 M
=1.1978 M
and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure ' also doubles from ,
Thus,Highest osmotic pressure that membrane may experience is, '=2*
=2*29.319 atm
' =58.638 atm
