# calculate the amount of KCL which must be added to 1Kg of water so that the freezing point is depressed by 2K.(Kf for water =1.86 K Kg/mol)?

Dear student!

We know that depression in freezing point,

ΔTf = Kf x m

Where m = molality i.e no. of moles of solute dissolved in per Kg solvent.

Here ,m = ΔTf /Kf =  2/1.86 = 1.07

We know that molality of KCl (m) = no. of moles of KCl /solvent in Kg.

So, for 1 Kg of water, no . of moles of KCl = 1.07 x 1= 1.07 moles.

Hence, 1.07 moles of KCl is dissolved in the solution.

• 21

acc to question, we have to find the mass of solute added in solvent.

therefore by formulae  ΔTf = Kf * Molarity

we are given  ΔTf = 2Kelvin And Kf as well

Put Values in formulae and u will get ur answer

THNX

• 7

Sorry , By mistake i Changed Molality INTO MOLARITY......SOrry for inconvenience And formulae for Molality Is

[Weight of solute Wb

----------------------------                                       X   1000

Molar Mass of solute * weight of solvent

• 4

thanks for help Sachin.

• -15
What are you looking for?