KCET · Chemistry · Solid State
A metal crystallises in bcc lattice with unit cell edge length of \(300 \mathrm{pm}\) and density \(6.15 \mathrm{gcm}^{-3}\). The molar mass of the metal is
- A \(50 \mathrm{gmol}^{-1}\)
- B \(60 \mathrm{gmol}^{-1}\)
- C \(40 \mathrm{gmol}^{-1}\)
- D \(70 \mathrm{gmol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(50 \mathrm{gmol}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, metal crystallises in bcc lattice, therefore \(Z=2\)
\(\begin{aligned} \text { Edge length } &=300 \mathrm{pm} \\ &=300 \times 10^{-10} \mathrm{~cm} \end{aligned}\)
Density, \(\begin{aligned} d &=\frac{Z M}{a^{3} N_{A}} \Rightarrow M=\frac{d a^{3} N_{A}}{Z} \\ &=\frac{6.15 \times\left(300 \times 10^{-10}\right)^{3} \times 6 \times 10^{23}}{2} \\ &-498150000 \times 10^{-7} \\ &=49.82 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \cong 50 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}\)
\(\begin{aligned} \text { Edge length } &=300 \mathrm{pm} \\ &=300 \times 10^{-10} \mathrm{~cm} \end{aligned}\)
Density, \(\begin{aligned} d &=\frac{Z M}{a^{3} N_{A}} \Rightarrow M=\frac{d a^{3} N_{A}}{Z} \\ &=\frac{6.15 \times\left(300 \times 10^{-10}\right)^{3} \times 6 \times 10^{23}}{2} \\ &-498150000 \times 10^{-7} \\ &=49.82 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \cong 50 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}\)
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