Question

5% of anhydrous cacl2 at 0°c developed 15atm osmotic pressure. what is degree of dissociation of cacl2

Answer 1

Answer:

Degree of dissociation = 0.2433 or 24.33%

Given:

The mass of $CaCl_{2}$ = 5g

To find:

Degree of dissociation

Explanation:

5g of $CaCl_{2}$ are present n 100ml, so 111g (mol. wt. of $CaCl_{2}$) will be present in

$\frac{100*111}{5*1000} = 2.22lt$

Now πV=ST(∵n=1)

$\pi =\frac{0.082*273}{2.22}=\frac{22.47}{2.22}=10.09 atm$

$i=\frac{Actually no. of particles in solution}{No. of particles taken}$  $and$  $\alpha =\frac{i-1}{n-1}$

here n=3

$\alpha =\frac{\frac{15}{10.09}-1 }{3-1}= 0.2433$

Important points about the degree of dissociation,

• Dissociation in chemistry is defined as breaking down a complex molecule, ionic species, or compound into simpler ions, atoms, or molecules.
• Dissociation is defined as the fraction of dissolved ions (current-carrying) that dissociate at a given temperature.
• Strong acids and strong bases have a degree of dissociation of 1.
• Increasing the temperature increases the degree of electrolyte dissociation.
• degrees of dissociation are represented by symbols $\alpha$

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