MENU
Question -

Calculate┬а(a)┬аmolality┬а(b)┬аmolarityand┬а(c)┬аmole fraction of KI if the density of 20% (mass/mass)aqueous KI is 1.202 g mL-1.



Answer -

(a)┬аMolarmass of KI = 39 + 127 = 166 g molтИТ1

20% (mass/mass)aqueous┬аsolution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 тИТ20)┬аg of water = 80 g of water

Therefore,molality of the solution

= 1.506 m

= 1.51 m (approximately)

(b)┬аItis given that the density of the solution = 1.202 g mLтИТ1

тИ┤Volume of 100 gsolution┬а

= 83.19 mL

= 83.19 ├Ч 10тИТ3┬аL

Therefore,molarity of the solution┬а

= 1.45 M

(c)┬аMolesof KI┬а
Moles of water
Therefore, mole fraction of KI

= 0.0263

Comment(S)

Show all Coment

Leave a Comment

Free - Previous Years Question Papers
Any questions? Ask us!
×