# 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) $1.51$

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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