Question

Calculate the final concentration (in Molarity) when 25 mL of water is added to 95 mL of 1.7 M LiCl. Show your work.

Part E - How many grams of KCl must be added to depress the freezing point of 1.00 kg of water to a temperature of -2.0 C? Show your work.

Part F - What is the maximum number of grams of KI that can be added to 40. g of water before any precipitation is formed at 20 C if the solubility is 140 g KI/ 100 g of H₂O? Show your work.

Answer 1

According to law of dilution   MV = M'V'

Where M = Molarity of stock = 1.7 M

V = Volume of the stock = 95 mL

M' = Molarity of dilute solution = ?

V' = Volume of the dilute solution = 95+25=120 mL

Plug the values we get   , M' = MV /V' = 1.35 M

Part E:

We know that ΔT _f = iK_f x m
Where

ΔT _f = depression in freezing point

        = freezing point of pure solvent – freezing point of solution

        = 0-( -2.0)

= 2.0 oC

K _f = depression in freezing constant of water = 1.86 oc/m

i= vanthoff’s factor = 2 ( since two ions K+ & Cl- )

m = molality of the solution

    = ( mass / Molar mass ) / weight of the solvent in Kg

= ( m/ 74.5) / 1.00 kg

= m/74.5

Plug the values we get 2.0 = 2x1.86x(m/74.5)

m = 40.0 g

Part F :

Given solubility is 140 g KI/ 100 g of H2O

that means 100 g of water contains 140 g of KI

40 g of water contains M g of KI

M = ( 40x140)/100

= 56 g