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