0.585%nacl solution at 27degree has osmotic pressure
Hello,
0.585% NACL solution at 27 degrees has osmotic pressure,
Osmotic pressure = i*C*R*T
where,
i = Van’t-Hoff factor (1 for non-electrolyte)
C = concentration of solute (in terms of Molarity)
R = Gas constant = 0.082 L (atm) (mol)^-1 K^-1
T = temperature in kelvin
0.585% NaCl solution means 0.585g Nacl is present in 100ml of solution.
mole of NaCL = weight given/Molecular weight of NaCl
mole of NaCL = 0.585g /58gmol^-1 = 0.01
Hence Concentration(C) of NaCl (in terms of Molarity)
= (mole of NaCl(n)/Volume of solution)*1000
= (0.01) / 100 *1000
= 0.10
Hence Osmotic pressure = 1∗(0.10)∗0.082∗300 atm = 2.46 atm
0.585% NACL solution at 27 degrees has osmotic pressure 2.46 atm.
Hope this answer will help you
Good luck
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