Question

Determine the amount of Calcium Chloride dissolved in 2.5 litre of water that it's osmotic pressure is 0.75 atm. temperature is 27°C

Vann Hoff factor is 2.47​

Answer 1

Explanation:

Solution

Given that

Osmotic pressure π = 0.75 atm

Volume V = 2.5 liter

van’t Hoff factor i = 2.47

T = 27 + 273 = 300 K

Let w is the amount of CaCl2 required

Gas constant R = 0.0821 L atm K-1mol-1

Molar mass of CaCl2M = 1 × 40 + 2 × 35.5 = 111g mol-1

Use the formula of osmotic pressure

Plug the values in this formula we get

W = 3.43 g

Answer

Required amount of CaCl2 = 3.42 g.

Answer 2

${\bold{\underline{Hey\:Mate!}}}$

Here's the Answer❗

${\bold{\boxed{\mathfrak{\underline{Colugative\:Properties}}}}}$

♦Given that ,

→ Volume of Water (V) = 2.5 l

→Osmotic pressure (π ) = 0.75 atm

→ Temperature (T) = 27°C

Using equation ,

π = i.CRT

π = i$\frac{n}{v}$RT

Now , let number of moles n = x

R = 0.082

$= > 0.75 = 2.47 \times \frac{x}{2.5} \times 0.082 \times 27 \\ \\ = > x = \frac{0.75 \times 2.5}{2.47 \times 0.082 \times 27} \\ \\ = > x = 0.03$

Let's consider the given mass be y g.

$n = \frac{given \: mass}{molecular \: mass} \\ \\ 0.03 = \frac{y}{111}$

•°• y = 111 × 0.03 = 3.33

→ ${\bold{\underline{\underline{The\:mass\:required \:is\:3.33g}}}}$

______________________________________________