UPSEE Solved Model Paper 7

Given: Molecular weight = 200 g/mole Edge length = $\overset{0}{3A}$ Avogadro's number = 6 × $10^{23} \mathrm{mol}^{-1}$ We have to find the density of the element The density of an element is given by the formula ρ = $\frac{Z \times M}{a^3 \times N_0}$ ... (i) where, Z = Number of atoms per unit cell (for simple cubic unit cell, Z = 1) M = Molecular weight a = Edge length N。= Avogadro's number = 6 × $10^{23} \mathrm{mol}^{-1}$ Now, a = $\overset{0}{3A}$ = 3 × $10^{-8}$ cm (Since $\overset{0}{1A}$ = $10^{-8}$ cm) Substituting all the values in equation (i). we get ρ = $\frac{1 \times 200}{(3 \times 10^{-8})^3 \times 6 \times 10^{23}}$ ρ = $\frac{200}{27 \times 6 \times 10^{-1}} \mathrm{g/cm}^3$ ρ = $\frac{2000}{27 \times 6} \mathrm{g/cm}^3$ ρ = 12.345 $\mathrm{g/cm}^3$ ~ 12.29 $\mathrm{g/cm}^3$ Therefore, the density of the element is 12.29 $\mathrm{g/cm}^3$. Hence, option 'C' is correct