Calculate molality of 2.5 g of ethanoic acid $[C H_{3} C O O H]$ in 75 g of benzene.

0.551 m o l k g^{- 1}

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

0.556 m o l k g^{- 1}

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

0.558 m o l k g^{- 1}

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

0.552 m o l k g^{- 1}

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

Open in App

Solution

The correct option is B $0.556 m o l k g^{- 1}$

MW of $C H_{3} C O O H = 60 g m o l^{- 1}$

$m = \frac{w_{2} \times 1000}{M w_{2} \times w_{1}} = \frac{2.5 g \times 1000}{60 g m o l^{- 1} \times 75 g} = 0.556 m o l k g^{- 1}$

Suggest Corrections

Similar questions

[C H_{3} C O O H]

(C H_{3} C O O H)

[C H_{3} C O O H]

Join BYJU'S Learning Program