Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739
For Business Enquiry

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

CBSEXII| CHEMISTRYMOST IMPORTANT QUESTIONSwww.topperlearning.com2CBSEClass XII ChemistryMost Important QuestionsChapter 1: The Solid State1.What is the formula of a compound in which the element Y forms hcp lattice and atoms of X occupy 2/3rdof tetrahedral voids?[1M]2.Givean example each of a molecular solid and an ionic solid.[1M]3.An element with density 2.8gcm-3forms a fccunit cellwith edgelength 4 × l0-8cm. Calculate the molar mass of the element.[2M](Given: NA=6.022 × 1023mol-1)4.[2M](i)What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?(ii)What type of stoichiometric defect is shown by NaCl? 5.How will you distinguish between the following pairs of terms?[2M](i)Tetrahedral and octahedral voids(ii)Crystal lattice and unit cell 6.Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 315.5pm, what is the radius of tungsten atom? [3M]7.Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm,the density of iron is 7.874 gcm-3. Use this information to calculate Avogadro's number. (At.Mass of Fe = 55.845 μ).[3M]8.An element with density 10 gcm−3forms a cubic unit cell with edge length of 3 × 10−8cm. What is the nature of the cubic unit cell if the atomic mass of the element is 81 gmol−1? [3M]

CBSEXII| CHEMISTRYMOST IMPORTANT QUESTIONSwww.topperlearning.com39.(a)An element has a body centered cubic structure with a cell edge of 288 pm. Thedensity of the elements is 7.2 gcm-3. Calculate the number of atoms present is 208 g of the element.[3M](b) With example explain difference between crystalline and amorphous solids. [2M]Chapter 2: Solutions1.A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling

Asked by Harsh.1102.raj 10th January 2019, 9:23 AM
Answered by Expert
Answer:

Q.1 What is the formula of a compound in which the element Y forms hcp lattice and atoms of XE occupy 2/3rd of tetrahedral voids?

 

     Element Y forms hcp lattice,

    No. of atoms of Y in hcp = 6

    No. of tetrahedral voids, X = 2 × 6 = 12

    Only 2/3rd of these are ocuppied by X so,

    12×2/3= 8

   So, X = 8 and Y = 6

    So the formula of the compound formed is X4Y3.

 

 

Q.2 An element with density 2.8g cm-3 forms a fcc unit cell with edge length 4 × l0-8cm. Calculate the molar mass of the element.[2M](Given: NA=6.022 × 1023 mol-1)

      Edge length a = 4 × 10-8 cm

     Density d, = 2.8 g cm-3

    As the lattice is fcc type, the no. of atoms per unit cell, z = 4

   NA=6.022 × 1023 mol-1 

 

    straight d space equals space fraction numerator straight z cross times straight m over denominator straight N subscript straight A space cross times space open parentheses straight a close parentheses cubed end fraction

space straight m space equals fraction numerator straight d cross times space straight N subscript straight A space cross times space open parentheses straight a close parentheses cubed over denominator straight z end fraction

space space space space space equals space fraction numerator 2.8 cross times 6.022 space cross times space 10 to the power of 23 cross times space open parentheses 4 space cross times space 10 to the power of negative 8 end exponent space close parentheses cubed over denominator space 4 end fraction

space space space space space equals space 26.97 space straight g divided by mol

space space space space space space almost equal to space 27 space straight g divided by mol

 

The molar mass of the element is 27 g/mol.
Answered by Expert 11th January 2019, 6:04 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp