Mathematics Physics Chemistry Videos Login/Register Ask Question

24 - Solid State Questions Answers

Chemistry >> Physical Chemistry >> Solid State Medical Exam

IF D POSITn OF Na+ & Cl - rR INTERCHANGED IN NaCl having FCC ARRANGEMENT OF Cl- THEN IN D UNIT CELL OF NaCl

Na + WILL INCREASE  BY 1 WHILE Cl- DECREASE BY 1

PLS TELL D REASON

Asked By: SARIKA
is this question helpfull: 4 0 submit your answer
Joshi sir comment

No, 

These will be no change by making this type of change

Chemistry >> Physical Chemistry >> Solid State Medical Exam

THE PYKNOMETRIC DENSITY OF NaClis 2.165X 10-3 wHILE ITS Xray density  IS 2.178X103 kgm-3 the fraction of unoccupied sites in NaCl crystal is?ans) 5.96x 10-3.   how we will caculate this fraction

Asked By: SARIKA
is this question helpfull: 5 0 submit your answer
Joshi sir comment

Pyknometer is a device to measure density of substance. Density measured by pyknometer is called pyknometric density. In porous solids this density may be different from true density of solid if the liquid used is unable to fill the pores of solid.

I think that the first density will be 2.165 * 103

difference of density = 2178-2165 = 13 kg/m3

so fraction of unoccupied site = 13/2178 =  5.96 * 10-3

Chemistry >> Physical Chemistry >> Solid State Medical Exam

CRYSTALLINE  SOLIDS HAVE

a) SHORT RANGE  ORDER

b) ANISOTROPIC DISTRIBUTn

c) LONG RANGE ORDER

D)  NO ORDER

AIIMS2001

SIR AS WE KNOW THAT BOTH THE OPTION B) & C) ARE CORRECT BUT IN THIS QUES WHICH OPTn WOULD BE MORE APPROPRIATE FOR CORRET ANS

 

 

 

 

 

 

 

 

 

 

 

 

 

.

Asked By: SARIKA
is this question helpfull: 2 0 submit your answer
Joshi sir comment

According to me it should be b

for anisotropic character is a basic property of crystalline solids

Chemistry >> Physical Chemistry >> Solid State Medical Exam

Q In NAcL type str of A+ &B- ,2A ATOMS ARE REMOVED FROM EDGES &2B ATOMS ARE REMOVED FROM CORNER. FORMULA=?  ANS =A14B15 . HOW?

Asked By: SARIKA
is this question helpfull: 2 1 submit your answer
Joshi sir comment

In NaCl like structure B covers all the corners and face centre and A covers all the edge centres and body centre

Net contribution of B = 8*1/8 + 6*1/2 = 4

Net contribution of A = 12*1/4 + 1*1 = 4

now 2 A ATOMS ARE REMOVED FROM EDGES so remaining contribution = 10*1/4 + 1*1 = 7/2

and 2 B ATOMS ARE REMOVED FROM CORNE so remainnig contribution = 6*1/8 + 6*1/2 = 15/4

so formula will be A7/2B15/4  or A14B15

Login Here

Register