Neet Solved Paper 2013 Question 12
Question: A metal has a fee lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g $ c{m^{-3}} $ . The molar mass of the metal is ( $ N _{A} $ Avogadros constant $ =6.02\times 10^{23}mo{l^{-1}} $ )
Options:
A) $ 40,g,mo{l^{\text{-1}}} $
B) $ 30,g,mo{l^{\text{-1}}} $
C) $ 27,g,mo{l^{\text{-1}}} $
D) $ 20,g,mo{l^{\text{-1}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given, cell is fee, so Z = 4 Edge length, a = 404 pm = 4.04 x 10-8 cm Density of metal,
$ d=2.72,g,c{m^{-3}} $ $ N _{A}=6.02\times 10^{23}mo{l^{-1}} $
Molar mass of the metal, M = ?
We know that density, $ d=\frac{Z\times M}{a^{3}\cdot N _{A}} $
$ \therefore $ $ M=\frac{d\cdot a^{3}\cdot N _{A}}{Z} $ $ =\frac{2.72\times {{(4.04\times {10^{-8}})}^{3}}\times 6.02\times 10^{23}}{4} $ $ =27,g,mo{l^{-1}} $
