ilver crystallises in fcc lattice, calculate the atomic mass of silver Chapter 1: the Solid State Chemistry Class 12 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. Question 11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10-8 cm and density is 0.5 g cm3, calculate the atomic mass of silver. is solved by our expert teachers. You can get ncert solutions and notes for class 12 chapter 1 absolutely free. NCERT Solutions for class 12 Chemistry Chapter 1: the Solid State is very essencial for getting good marks in CBSE Board examinations
Question 11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10-8 cm and density is 0.5 g cm3, calculate the atomic mass of silver.
Answer
Edge of length of cell a = 4.07x10-8cm
Density p = 10.5 g /cm3
Number of atoms in unit cell of fcc lattice = 4
Avogadro number NA = 6.022x1023
Use formula
Density $p = /frac{{ZM}}{{{a^3}{N_A}}}$
Cross multiply we get
ZM = pa3NA
Divide by Z we get
/[M = /frac{{p{a^3}{N_A}}}{Z}/]
Plug the value we get
/[M=/frac{10.5{{(4.07)}^{3}}6.022x{{10}^{23}}}{4}/]
/[M = /frac{{10.5x67.41x6.022}}{4}/]
/[M = 107.09g{/rm{ }}mo{l^{ - 1}}/]
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