Question

the value of van't hoff factor 'i' for CaCl2 assumingcomplete dissociation of solute is​

Answer 1

Answer:

Hope it helps you!

Explanation:

The osmotic pressure (calculate), π=

V

n

RT....(i)

where, V=volume=2.5 L,T=Temperature=300K,n= dissolving moles of

CaCl

2

=

molecular mass of solute (M)

wt of solute (g)

Thus equation (i) becomes π

Cal

=

MV

8

RT

∴π

Cal

=

111g mol

−1

×2.5l

g

×0.082 l atm K

−1

mol

−1

×300K

=

277.5

24.6 g

atm=0.0886×g.atm.gm

Van't Hoff factor (i) =

calculated osmotic pressure(π

cal

)

observed osmotic pressure(π

ob

)

∴2.47=

π

Cal

0.75 atm

or, π

Cal

=0.3036 atm

Thus, 0.3036=0.0886×g gm or,

g=3.4271 gm.

Hence, the amount of CaCl

2

dissolve in 2.71 solution is 3.4271 gm.