0.585% NaCl solution at 300K has osmotic pressure?

1 Answer
Nov 21, 2017

#Π = "5.23 bar"#

Explanation:

The osmotic pressure of a salt solution is given by the formula

#color(blue)(bar(ul(|color(white)(a/a)Π = icRTcolor(white)(a/a)|)))" "#

where

#i = # the van't Hoff #i# factor for the solute
#c =# the molar concentration of the solute
#R =# the universal gas constant
#T=# the Kelvin temperature

Our first task is to calculate the molar concentration of the #"NaCl"#.

0·585 % (m/m) #"NaCl"# means that we have 0.585 g #"NaCl"# in 100 g solution

The solution is so dilute that we can assume the density is the same as that of water.

Thus, we have 0.585 g #"NaCl"# in 100 mL of solution.

#c = (0.585 color(red)(cancel(color(black)("g NaCl"))))/(100 color(red)(cancel(color(black)("mL solution")))) × (1000 color(red)(cancel(color(black)("mL solution"))))/("1 L solution") × ("1 mol NaCl")/(55.84 color(red)(cancel(color(black)("g NaCl")))) = "0.1048 mol/L"#

Also, #i = 2# for #"NaCl"#.

Hence,

#Π = icRT = 2 ×0.01048 color(red)(cancel(color(black)("mol·L"^"-1"))) ×"0.083 14 bar"·color(red)(cancel(color(black)("L·K"^"-1""mol"^"-1"))) × 300 color(red)(cancel(color(black)("K"))) = "5.23 bar"#