Heat of combustion of benzene is 718 K.cals. When 39 gms of benzene undergoes combustion, the heat liberated is _______.

718 K.cals

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

359 K.cals

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

135 K.cals

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

1436 K.cals

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

Open in App

Solution

The correct option is B 359 K.cals
moles $= \frac{39}{78} = 0.5$
$= \frac{h e a t l i b e r a t e d}{h e a t o f c o m b u s t i o n} =$ moles $= \frac{1}{2}$

$∴$ heat liberated $= \frac{1}{2}$ x heat of combustion
$= \frac{1}{2}$ x 718
$=$ 359 K.cals

Suggest Corrections

Similar questions

27^{\circ} C

7.8 g

327 k J

27^{o} C

- 3264 k J m o l^{- 1}

39 g

Join BYJU'S Learning Program