1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
When 1 mol of Znis dissolved in excess HCI the work done is approximately equal to -2.46 kl in open beaker at 300 K and 1 atm.
b
When 1 mol of Zn is dissolved in excess HCI work done is equal to zero in closed beaker
c
Both (1) and (2) are correct
d
Neither (1) nor (2) is correct.
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Zn+2HCl→ZnCl2+H2The no. of moles of H2 produced = 1 mol Vol. of H2(g) produced =0.082×3001L=24.6L=24.6×10−3m3∴ in open beaker, ΔW=−Pext×ΔV=−105×24.6×10−3=−2460J In a closed vessel, ΔV=0 and ΔW=0