Question:
For two isomorphous crystals A and B, the ratio of density of A to that of B is 1.6 while the ratio of the edge length of B to that of A is 2. If the molar mass of crystal B is 200 g mol$^{-1}$, then molar mass of crystal A is}
For two isomorphous crystals A and B, the ratio of density of A to that of B is 1.6 while the ratio of the edge length of B to that of A is 2. If the molar mass of crystal B is 200 g mol$^{-1}$, then molar mass of crystal A is}
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Density varies inversely with cube of edge length.
Updated On: May 8, 2026
- 240 g mol$^{-1}$
- 120 g mol$^{-1}$
- 80 g mol$^{-1}$
- 160 g mol$^{-1}$
- 40 g mol$^{-1}$
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The Correct Option is
Solution and Explanation
Concept:
\[
\rho \propto \frac{M}{a^3}
\]
Step 1: Write ratio. \[ \frac{\rho_A}{\rho_B} = \frac{M_A/a_A^3}{M_B/a_B^3} \]
Step 2: Substitute values. \[ 1.6 = \frac{M_A}{M_B} \cdot \frac{a_B^3}{a_A^3} \] Given: \[ \frac{a_B}{a_A} = 2 \Rightarrow \frac{a_B^3}{a_A^3} = 8 \]
Step 3: Substitute. \[ 1.6 = \frac{M_A}{200} \times 8 \]
Step 4: Solve. \[ \frac{M_A}{200} = \frac{1.6}{8} = 0.2 \] \[ M_A = 40 \]
Step 5: Final answer. \[ \boxed{40\,g\,mol^{-1}} \]
Step 1: Write ratio. \[ \frac{\rho_A}{\rho_B} = \frac{M_A/a_A^3}{M_B/a_B^3} \]
Step 2: Substitute values. \[ 1.6 = \frac{M_A}{M_B} \cdot \frac{a_B^3}{a_A^3} \] Given: \[ \frac{a_B}{a_A} = 2 \Rightarrow \frac{a_B^3}{a_A^3} = 8 \]
Step 3: Substitute. \[ 1.6 = \frac{M_A}{200} \times 8 \]
Step 4: Solve. \[ \frac{M_A}{200} = \frac{1.6}{8} = 0.2 \] \[ M_A = 40 \]
Step 5: Final answer. \[ \boxed{40\,g\,mol^{-1}} \]
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