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An element with density 6 g cm-3 forms a fcc lattice with edge length of 4 x 10-8 cm. The molar mass of the element is (NA = 6 x 1023 mol-1) :(A) 57.6 g mol-1(B) 28.8 g mol-1(C) 82.6 g mol-1(D) 62 g mol-1 |
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Answer» Option : (A) 57.6 g mol-1 We know that, Number of atoms per unit cell of FCC = 4 Z = 4 We have given - d = 6g/cm3, a = edge length = 4 x 10-8 cm NA = 6 x 1023 d = \(\frac{ZM}{a^3N_A}\) M = molar mass of element. ∴ 6g/cm3 = \(\frac{4\times M}{(4\times 10^{-8})^3cm^3\times 6\times 10^{23}mol^{-1}}\) ⇒ 6 x (64 x 10-24) x 6 x 1023 = 4M ⇒ M = \(\frac{6\times 6\times 64\times 10^{-1}}{4}g/mol\) ∴ M = 57.6 g/mol. |
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