$2 m o l e$ of zinc is dissolved in $H C l$ at $25^{o} C$ . The work done in open vessel is :

- 2.477 k J

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

- 4.955 k J

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

0.0489 k J

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

N o n e

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

Open in App

Solution

The correct option is C $- 4.955 k J$
Given $n = 2; T = 25^{o} C = 298 K; R = 8.314 J / m o l . K$
$W$ is the work done.
$W = - P △ V$
$= - n R T$
$W = - 2 \times 8.314 \times 298$
$= - 4.955 k J$
Hence,option B is correct.

Suggest Corrections

Similar questions

2

H C l

25^{o} C

2 mol of zinc is dissolved in $H C l$ at $25^{\circ} C$ . The work done in an open vessel is:

2 mole of zinc is dissolved in HCl at $25^{\circ} C$ . The work done in open vessel is:

2 mol of zinc is dissolved in $H C l$ at $25^{\circ} C$ . The work done in an open vessel is:

11.2 g

H C l

25^{o} C

1

Join BYJU'S Learning Program