$20$ g of $N a O H$ (Molar mass $= 40$ g $m o l^{- 1}$ ) is dissolved in $500$ $c m^{3}$ of water. Molality of resulting solution is:

0.1

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

0.5

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

1.5

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

1.0

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

Open in App

Solution

The correct option is B $1.0$ m
We know,

Molality $= \frac{m o l e s}{w e i g h t o f s o l v e n t (k g)}$

Molality $= (\frac{w e i g h t}{m o l e c u l a r w e i g h t}) \times \frac{1000}{(v o l u m e \times d e n s i t y)}$ [Note: density $= 1 g m / c m^{3}$ ]

Molality $= \frac{20}{40} \times \frac{1000}{500} \times 1 = 1 m$

Suggest Corrections

Similar questions

g m o l^{- 1})

N a O H

40 g {m o l}^{- 1})

1.6 g

N a O H

500 c m^{3}

N a O H

K N O_{3}

K N O_{3}

K N O_{3}

Join BYJU'S Learning Program