Question #da8a3

1 Answer
Feb 28, 2016

Answer:

The molarities are a) 0.0279 mol/L and b) 0.0186 mol/L.

Explanation:

The van't Hoff factor #i# is the number of moles of solute particles obtained from 1 mol of solute.

#"NaCl" → "Na"^+ + "Cl"^"-"#

For #"NaCl", i = 2#, because 1 mol of #"NaCl"# produces 2 mol of ions, and each ion contributes equally to the osmotic pressure.

#"Na"_2"SO"_4 → "2Na"^+ + "SO"_4^"2-"#

For #"Na"_2"SO"_4, i = 3#, because 1 mol of #"Na"_2"SO"_4# produces 3 mol of ions, and each ion contributes equally to the osmotic pressure.

a) #"NaCl"#

#Π = iMRT#

#M = Π/(iRT) = (1.25 color(red)(cancel(color(black)("atm"))))/("2 × 0.08 206"color(white)(l) "L·"color(red)(cancel(color(black)("atm""·K"^"-1")))"mol"^"-1" × 273.15color(red)(cancel(color(black)("K"))))= "0.0279 mol/L"#

b) #"Na"_2"SO"_4#

#M = Π/(iRT) = (1.25 color(red)(cancel(color(black)("atm"))))/("3 × 0.082 06 L"·color(red)(cancel(color(black)("atm·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))= "0.0186 mol/L"#