Question

What is the osmotic pressure of 0.30M solution of MgSO4 if the MgSO4 is 80% dissociated at 20°C (i) Vant Hoff factor (given R=0.0821 LatmK-1 mol-1 (i)

Answer 1

The osmotic pressure of given amount of magnesium sulfate is 12.98 atm

Explanation:

The relationship between Van't Hoff factor and degree of dissociation is given as:

$\alpha=\frac{i-1}{n-1}\times 100$

where,

n = number of ions = 2  (for magnesium sulfate)

i = Van't Hoff factor = ?

$\alpha$ = degree of dissociation = 80 % = 0.80

Putting values in above equation, we get:

$0.8=\frac{i-1}{2-1}\times 100\\\\i=1.8$

To calculate the osmotic pressure, we use the equation:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 1.8

M = molarity of solute = 0.30 M

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $20^oC=[273+20]=293K$

Putting values in above equation, we get:

$\pi=1.8\times 0.30\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 293K\\\\\pi=12.98atm$

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