Question

Question #da8a3

Answer 1

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"`