Question da8a3

Feb 28, 2016

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.

$\text{NaCl" → "Na"^+ + "Cl"^"-}$

For $\text{NaCl} , i = 2$, because 1 mol of $\text{NaCl}$ produces 2 mol of ions, and each ion contributes equally to the osmotic pressure.

$\text{Na"_2"SO"_4 → "2Na"^+ + "SO"_4^"2-}$

For ${\text{Na"_2"SO}}_{4} , i = 3$, because 1 mol of ${\text{Na"_2"SO}}_{4}$ produces 3 mol of ions, and each ion contributes equally to the osmotic pressure.

a) $\text{NaCl}$

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) ${\text{Na"_2"SO}}_{4}$

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"#