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Question

Calculate:

A: Molality of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL1.

B: Molarity of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL1.

C: Mole Fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL1.

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Solution

A:

Given:

20% (mass/mass) aqueous solution of KI.
This means 20 g of KI is dissolved in 80 g of water.
So, mass of solute W2=20g and mass of solvent W1=80
Molar mass of solute (KI)
=39+127
=166 g mol1

Molality of KI:

We know, molality(m)=mass of solute(g)molar mass ofsolute×mass of solvent(kg)

Density of the aqueous KI=1.202 g mL1

m=20×1000166×80=250166=1.5m

Final Answer:
The molality of KI is 1.5m.

B:

Given:
Mass percentage=20%
And density of solution =1.202 g mL1

Molarity of KI
We know, molarity (M)=mass percentage×density×10molar mass

Molarity(M)=20×1.202×10166=1.45M

Final Answer:
The molarity of KI is 1.45M.

C:

Given:
20% (mass/mass) aqueous solution of KI.
This means 20 gof KI is dissolved in 80 g of water.
So, mass of solute (W2)=20g and mass of solvent (W1)=80g
Molar mass of solute (KI)
=39+127
=166gmol1

Moles of compound

Number of moles of KI,

nKI=Mass of KIMolar mass of KI=20166=0.12 mol

Number of moles of H2O,

nH2O=Mass of H2OMolar mass ofH2O=8018=4.44 mol

Mole fraction of compounds

Now, mole fraction of (KI),

χKI=nKInKI+nH2O

χKI=0.120.12+4.44
(χKI)=0.124.56=0.0263

Final Answer:
The mole fraction of KI in the given solution is 0.0263.

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