Assuming that sea water is a 3.5 wt % solution of nacl in water, calculate its osmotic pressure at 20°c. the density of a 3.5% nacl solution at 20°c is 1.023 g/ml.

π = 14.824 atm

Explanation:

wt % = ( w NaCL / w sea water ) * 100 = 3.5 %

assuming w sea water = 100 g = 0.1 Kg

⇒ w NaCl = 3.5 g

osmotic pressure ( π ):

∴ T = 20 °C  + 273 = 293 K

∴ C ≡ mol/L

∴ density sea water = 1.03 Kg/Lfrom literature

⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln

⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol

⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M

⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K

⇒ π = 14.824 atm

answer; ///i do believe that the correct answer is (b); ///particle composed of two or more atoms bonded together;

am i too late?

explanation: