Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
Calculate the molar mass of an element having density $21 \mathrm{~g} \mathrm{~cm}^{-3}$ that forms fcc unit cell $\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=36 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]$
ChemistrySolid StateMHT CETMHT CET 2023 (09 May Shift 2)
Options:
  • A $\quad 292.00 \mathrm{~g} \mathrm{~mol}^{-1}$
  • B $189.00 \mathrm{~g} \mathrm{~mol}^{-1}$
  • C $140.00 \mathrm{~g} \mathrm{~mol}^{-1}$
  • D $108.00 \mathrm{~g} \mathrm{~mol}^{-1}$
Solution:
2457 Upvotes Verified Answer
The correct answer is: $189.00 \mathrm{~g} \mathrm{~mol}^{-1}$
$\begin{aligned} & \operatorname{Density}(\rho)=\frac{M \text { n }}{a^3 N_A} \\ & \therefore \quad 21 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{\mathrm{M} \times 4}{36 \mathrm{~cm}^3 \mathrm{~mol}^{-1}} \\ & \therefore \quad \mathrm{M}=\frac{21 \mathrm{~g} \mathrm{~cm}^{-3} \times 36 \mathrm{~cm}^3 \mathrm{~mol}^{-1}}{4} \\ & =189.00 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.