214 - Chemistry Questions Answers

IN A CUBIC CRYSTAL OF CsCl  (d= 3.97gm/cc) THE 8 CORNER OCCUPIED BY Cl-  WITH Cs+ AT CENTRE . CALCULATE THE RATIO OF RADIUS OF Cl- & Cs+

Asked By: SARIKA SHARMA
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Joshi sir comment

use the formula d = Mn/N0a3

calculate a?                here n = 1

now 2r - + 2r = a √3

and 2r = a

divide these equations for getting ratio

in a crystal A2B3 ATOMS B ARE IN CCP ARRANGMT. & ATOMS A ARE IN TETRAHEDRAL VOIDS . THE FRACTN OF TETRAHEDRAL VOIDS ARE OCCUPIED IS ----?

Asked By: SARIKA SHARMA Solved By: AMIT DAS
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Joshi sir comment

B atom forms ccp, let No. of atoms of B = 1

no. of tetrahedral voids = 2

but formula given is A2/3B

so % of voids occupied = (2/3)*100/2 = 33.34 %  

if all the ions of NaCl str is removed from one of the diagonal plane , then the formula of compd. will be ---- ?

Asked By: SARIKA SHARMA
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Solution by Joshi sir

Normal formula of compound is Na4Cl4

now on removing all the atoms from 1 body diagonal, 

lost atoms = 4 Cl at corner, 2 Na from edge centre, 2 Cl from face centre, 1 Na from body centre

so loss in Cl = 4*1/8 + 2*1/2 = 1.5

and loss iin Na = 2*1/4 + 1*1 = 1.5

so new formula = Na2.5Cl2.5

so formula will remail same 

In FCC unit cell distance b/w centre of two spheres which are touching each other & distance b/w centre of two nearest non touching sphere is

1)  (2a)1/2,, 2a

2)2r, (a+r)

3)  a/(21/2), a

4) both (2) & (3)

a= edge length. r= radius of atom

Asked By: SARIKA SHARMA
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Joshi sir comment

along diagonal of a face 4r = a√2   or  r = a√2/4 so distance b/w centre of two touching spheres is 2r = a√2/2

similarly distance b/w two nearst non touching sphere = a (along edge)

so 3) will be the option

 

in a cubic closed packing of mixed oxide, it is found that lattice has O-2 ion & one half of the octahedral voids are occupied by trivalent cation A+3 & ONE EIGHTH of the FORMULA OF COMPOUND IS

1) AB2O4

2)  A2B2O4

3) A2BO4

4)ABO4

Asked By: SARIKA SHARMA
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Joshi sir comment

the question is incomplete, after ONE EIGHTH, i guess, it will be one eighth of the tetrahedral void are occupied by B

so let no. of O = 1

then no. of octahedral voids = 1 and half of this = 1/2

similarly tetrahedral voids = 2 so 1/8 of it = 2*1/8 = 1/4

so formula = A.5B.25O1  or A2B1O4

if my assumption for question is correct then answer would be 3)     here B is divalent positive ion.

what are d cordination no. of NaCl, ZnS, CsCl, CaF2,,Li2O   upto four places, is there any trick to remember all these? 

Asked By: SARIKA SHARMA
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Joshi sir comment

NaCl    fcc structure   6 for Na and 6 for Cl

ZnS      In ZnS , S forms the face centered cube and Zn occupies half of the tetrahedral holes..The unit cell can be devided into eight equal small cubes,each with 1 Zn and 4 S ions so the coordination no will be 4 for Zn and 4 for S

CsCl  bcc structue  8 for Cs and 8 for Cl

CaF2   fcc structure of Ca with 100% filled tetrahedral voids by F, so coordination no. of Ca is 8 and that of F is 4 

Li2O  anti fluorite structure, fcc structure of O with 100% filled tetrahedral voids by Li so coordination of O is 8 and that of Li is 4 

 

 

100 ml OF AN IDEAL GAS IS HEATED IN AN OPEN VESSEL FROM 300K TO 400K . THE VOLUME OF GAS THAT WILL REMAIN IN THE VESSEL IS =?

ANS) 100ML

Asked By: SARIKA SHARMA
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Joshi sir comment

its simple because question is based on volume and gas covers the complete volume of the vessel in which it is placed.

AT WHAT TEMP. THE r.m.s  VELOCITY OF OXYGEN WILL BE AS THAT OF METHANE AT 270 C ?

1) 540C

2) 327K

3) 600K

4)573K

Asked By: SARIKA SHARMA
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Joshi sir comment

v = {3RT/M}1/2

so according to the given condition 

{3RT/32}1/2 = {3R300/16}1/2

or T/32 = 300/16

or T = 600

so t = 600-273 = 3270C

Q. 1.Calculate the number of HCl molecules present in 10ml of 0.1N HCl solution.

Q.2. 4.0g of NaOH are contained in one decilitre of solutions. Calculate mole fraction. Molarity and molality of NaOH. ( Given, density of solution=                                                              1.038g/ cm3)

Q. 3. You are provided with 2 litre each of 0.5 M NaOH and 0.4M of NaOH solutions. What maximum volume of 0.30 M NaOH can be obtained from these solutions without using water at all ?

 

Asked By: HEMA RANI
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Joshi sir comment

1)   no of miligm eq. = 10*0.1 = 1

so no. of gm. eq. = 1/1000

since acid is monobasic so no. of moles = 1/1000

so molecules = 1/1000 * 6.023 * 1023

 

2)     0.1 litre soln. contains              4 gm. NaOH

       so 1 litre contains                      40 gm. NaOH

       so 1 litre soln. contains              40/40 moles of NaOH  = 1 mole so molarity = 1

 

 now 1 litre soln. contains                       40 gm NaOH

so 1*1038 gm soln. contains               40 gm. NaOH                   here 1038 gm/litre is density

so 1038 gm. soln. contains                  40 gm. NaOH

so 1038-40 gm. solvent contains        40 gm. NaOH

so 998 gm. solvent contains                 40 gm. NaOH = 1 mole 

so 0.998 Kg solvent contains                 1 mole NaOH

so molality = 1/0.998 

 

998 gm water contains                   40 gm. NaOH

so 998/18 mole water contains    40/40 mole NaOH

so mole fraction of NaOH = [40/40]/{[40/40]+[998/18]}

 

Third part is irrelevent without using water

 

 

 

depression in freezing point of 0.01 molal acetic acid solution is 0.0204 deg.cel. 1molal of this solution freezes at -1.86deg.cel.assuming molality equals molarity find pH of acid A. 2 B. 3 C. 3.2 D. 4.2
Asked By: GAURAV MAHATE
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Joshi sir comment

for 1 molal soln. depression = 2.04 deg. cel.

and normal depression = 1.86

so vant hoff factor = 2.04/1.86 = 102/93 = 34/31

now CH3COOH --------->   CH3COO-  +  H+

let degree of dissociation = α                 

so concentrations after dissociation will be 

       1-α                                   α                  α

so according to these concentrations vant hoff factor = 1+α/1

on comparing 1+α = 34/31 or α = 3/31               so pH = -log3/31

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