Van't hoff factor for srcl2 at 0.01 m is 1.6. Percent dissociation of

1.	Van't hoff factor for srcl2 at 0.01 m is 1.6. Percent dissociation of srcl2 is
Answer» Answer: Using the equation given below: Where, i=van't hoff factor n=NUMBER of particles that dissociate =DEGREE of dissociation Dissociation of SrCl₂ , can be seen in the equation given below: SrCl₂----->Sr²⁺ + 2Cl⁻ So number of PATICLES dissociate=3 i=1.6 putting all the VALUES in the equation: =0.3 As is the degree of dissociation then PERCENT dissociation=0.3×100 =30 percent Read more on Brainly.in - brainly.in/question/5532050#readmore

Van't hoff factor for srcl2 at 0.01 m is 1.6. Percent dissociation of srcl2 is

Answer»

Answer:

Using the equation given below:

Where,

i=van't hoff factor

n=NUMBER of particles that dissociate

=DEGREE of dissociation

Dissociation of SrCl₂ , can be seen in the equation given below:

SrCl₂----->Sr²⁺ + 2Cl⁻

So number of PATICLES dissociate=3

i=1.6

putting all the VALUES in the equation:

=0.3

As is the degree of dissociation

then PERCENT dissociation=0.3×100

=30 percent