You visited us 1 times! Enjoying our articles? Unlock Full Access!

Mole Fraction

What would be...

Question

What would be the molality of $20 %$ (mass/mass) aqueous solution of $K I$ ?
Molar mass of $K I$ is $166 g m o l^{- 1}$

1.51

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.35

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.08

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.48

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1.51$
$20 %$ (mass/mass) aqueous solution of $K I$ means $20 g$ of $K I$ in $100 g$ of solution.
Mass of solution =100g
Mass of solute=20g
Mass of solvent=mass of solution - mass of solute
$= (100 - 20) g = 80 g = 0.08 k g$
Molar mass of $K I$ is $166 g m o l^{- 1}$
$Number of moles of solute (n) = \frac{mass of KI}{molar mass of KI} = \frac{20}{166} = 0.121 m o l$
So, molality of solution $m = \frac{number of moles of solute}{weight of solvent (in kg)}$
$\Rightarrow m = \frac{0.121}{0.08} = 1.51 m o l / k g$
Hence, option (A) is the correct.

Suggest Corrections

Similar questions

20 %

K I

K I

166 g m o l^{- 1}

A

K I

20 % (mass/mass)

K I

1.202 g m L^{- 1}

B

K I

20 % (mass/mass)

K I

1.202 g m L^{- 1}

C

K I

20 % (mass/mass)

K I

1.202 g m L^{- 1}

K I

20 %

K I

1.202 g m L^{- 1}

1.202 g m L^{- 1}

K_{2} C r_{2} O_{7}

254 g I_{2}

K I

K_{2} C r_{2} O_{7} + K I \to C r^{3 +} + I_{2}

K_{2} C r_{2} O_{7}

294 g m o l^{- 1}

Join BYJU'S Learning Program