11. what is the osmolarity of 2.0 l of an aqueous solution containing 20.5 g of na2so4?

Explanation:

Step1: Calculate molar mass of Na₂SO₄

The molar mass of Na is 22.99 g/mol, S is 32.07 g/mol and O is 16.00 g/mol. For Na₂SO₄, $M=(2\times22.99 + 32.07+4\times16.00)\text{ g/mol}=142.05\text{ g/mol}$

Step2: Calculate number of moles of Na₂SO₄

$n=\frac{m}{M}$, where $m = 20.5\text{ g}$ and $M = 142.05\text{ g/mol}$. So $n=\frac{20.5\text{ g}}{142.05\text{ g/mol}}\approx0.144\text{ mol}$

Step3: Determine the number of particles per formula unit

Na₂SO₄ dissociates into 2 Na⁺ and 1 SO₄²⁻, so $i = 3$ (van - 't Hoff factor)

Step4: Calculate molarity

$Molarity=\frac{n}{V}$, where $n = 0.144\text{ mol}$ and $V=2.0\text{ L}$. So $M=\frac{0.144\text{ mol}}{2.0\text{ L}} = 0.072\text{ M}$

Step5: Calculate osmolarity

$Osmolarity=i\times Molarity$, with $i = 3$ and $Molarity=0.072\text{ M}$. So $Osmolarity=3\times0.072\text{ osmol/L}=0.216\text{ osmol/L}$

Answer:

0.216 osmol/L