What would be the molality of 20\:^{o}/_{o}  (mass/mass) aqueous solution of KI?

( molar mass of KI =166\:g\:\:mol^{-1}  )

  • Option 1)

    1.08

  • Option 2)

    1.35

  • Option 3)

    1.48

  • Option 4)

    1.51

Answers (1)

 

Molality -

Molality (m) = (number of moles of solute)/(mass of solvent in kg)

 

- wherein

It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by m.

 

 

 

Mass percent of solution -

Mass percent = ((mass of solute)/(mass of solution) ) X 100

 

- wherein

 it is mass of solute present in 100 gram solution.

 

 

 

 

 

20\:^{o}/_{o} (mass/mass) = 20 gm is present in 100 gm solution .

solute is KI, weight =20 gm (given)

solvent weight is =  100 - 20 = 80 gm

\\mole\:\:of\:\:solute\:\:=\frac{20\:\:gm}{166\:\:gm/mole}=\frac{20}{166}moles \\\\\\\:molality =\frac{mole\:\:of \:\:solute\:\:(K.I.)}{weight\:\:of\:\:solvent(kg)}=\frac{20/166}{80\times10^{-3}}\\\\\\\:m=1.51\:\:mole/kg


Option 1)

1.08

Option 2)

1.35

Option 3)

1.48

Option 4)

1.51

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