The Question is from NEET guide , and it was created by the author A.RAMA NAGESWARARAO

Question

A hypothetical metallic element crystallizes in a face-centered cubic lattice with a density of $10.5 \ g/cm^3$ and an atomic radius of $1.34 \unicode{x212B}$. Assuming perfect close-packing, calculate the atomic mass of the element (in amu).

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Crystal Structure and Density
	**Related Topics**: Face-centered cubic lattice, Avogadro's number, Unit cell
	**Prerequisites**: Unit cell dimensions, Density formula, Atomic radius
## SOLUTION PATHWAY
**Given Information**:
	Density ($ho$) = $10.5 \ g/cm^3$
	Atomic radius (r) = $1.34 \unicode{x212B} = 1.34 	imes 10^{-8} \ cm$
	Face-centered cubic (fcc) lattice
**Method & Steps**:
	1. Calculate the edge length (a) of the unit cell:
		Reasoning: In an fcc lattice, the relationship between the edge length (a) and the atomic radius (r) is given by $a = 2\sqrt{2}r$.
		Formula/Rule: $a = 2\sqrt{2}r$
	2. Calculate the volume (V) of the unit cell:
		Working: $V = a^3$
	3. Determine the number of atoms (n) per unit cell:
		Reasoning: An fcc unit cell contains 4 atoms.
		Key Point: n = 4 for fcc
	4. Calculate the mass (m) of one unit cell:
		Reasoning: Mass is related to density and volume by $ho = \frac{m}{V}$.
		Working: $m = ho 	imes V$
	5. Calculate the mass of one atom:
		Working: Mass of one atom = $\frac{m}{n}$ 
	6. Convert the mass of one atom to amu:
		Reasoning:  Use Avogadro's number ($N_A = 6.022 	imes 10^{23} \ atoms/mol$) to convert the mass from grams per atom to grams per mole (which is numerically equal to amu).
		Working: Atomic mass (amu) = Mass of one atom (g/atom) $	imes N_A$ (atoms/mol)
		Result: Atomic mass $\approx 86.1 \ amu$
		Verification: The calculated atomic mass is close to the known atomic mass of elements like Strontium (Sr).
## OPTION ANALYSIS
[A] :x: 107.9 amu: Incorrect. This value might result from errors in calculating the unit cell edge length or using a wrong number of atoms per unit cell.
[B] ✓: 86.1 amu: Correct. This is the calculated atomic mass using the correct formula and values.
[C] :x: 63.5 amu: Incorrect. This value is significantly lower than the expected range and may indicate errors in unit conversions or calculations.
[D] :x: 126.9 amu: Incorrect. This value is higher than the correct answer and might result from using the wrong formula for edge length or misinterpreting the crystal structure.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrect formula for edge length: Ensure the correct relationship between 'a' and 'r' is used for the given crystal structure (fcc in this case).
		2. Wrong number of atoms per unit cell: Remember n=4 for fcc, n=2 for bcc, and n=1 for simple cubic.
	**Quick Tips**:
		1. Memorize the formulas relating edge length and atomic radius for different cubic structures.
		2.  Always double-check unit conversions and significant figures.

A.RAMA NAGESWARARAO · Answer

107.9 amu

A.RAMA NAGESWARARAO · Answer

The correct Option is "86.1 amu"

A.RAMA NAGESWARARAO · Answer

63.5 amu

A.RAMA NAGESWARARAO · Answer

126.9 amu

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