What amount of CaCl2 (i = 2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5 atm at 27 °C?

Question

What amount of CaCl2 (i = 2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5 atm at 27 °C? (A) 3.42 g (B) 9.24 g (C) 2.834 g (D) 1.820 g

Tardigrade Admin · Accepted Answer

π = iCRT = i (n/V) RT n = (π × V/i× R× T) = (0.5 × 2/2.47 × 0.0821 × 300) = 0.0164 mol Amount of CaCl2 = n × M= 0.0164 × 111 = 1.820 g

Q. What amount of $C a C l_{2} (i = 2.47)$ is dissolved in $2$ litres of water so that its osmotic pressure is $0.5 a t m$ at $27^{°} C$ ?

$3.42 g$

$9.24 g$

$2.834 g$

$1.820 g$

$π = i CRT = i \frac{n}{V} RT$
$n = \frac{π \times V}{i \times R \times T}$
$= \frac{0.5 \times 2}{2.47 \times 0.0821 \times 300} = 0.0164 m o l$
Amount of $C a C l_{2} = n \times M = 0.0164 \times 111 = 1.820 g$

Q. What amount of CaCl2​(i=2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5atm at 27°C?

3.42g

9.24g

2.834g

1.820g