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5. Gold metal crystallizes in a face -centred cubic
unit cell (fcc). Determine the density of gold.
(Atomic mass of gold =179 u, atomic radius =
0.144nm,
| N = 6.022x10 23 mol-' )
(গ’ল্ড ধাতুৱে পৃষ্ঠকেন্দ্রিক ঘনকীয় (fcc) একক কোষ গঠন
কৰে। গ’ল্ডৰ ঘনত্ব নির্ণয় কৰা। (গ’ল্ডৰ পাৰমাণৱিক ভৰ |
=179 u, পাৰমাণৱিক ব্যাসার্ধ =0.144nm,
Answers
Answer:
✨ ✨༺ H E Y 彡M A T E ࿐✨ ✨
Here Is Your Answer, Hope It helps
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Answer:-Here Z= 4 (since it's FCC)
M= 197
r = 0.144
a = ?
Therefore, Step-I. Calculate of edge length of unit cell.
a=2√2r
a=2√2 × 0.144 nm
a=407.23 pm (unit converted)
Step-II. Calculation of density of unit cell.
Density=Z×M/a^3×No×10^-30
Density =4×197/(407.23) ^3 × 6.022 × 10^23 × 10^-30
Density= 19.38 g/cm^3
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Extra - Here,
(No)is Avogadro Number
(a^3) is a cube [for clarification so u won't have doubt ]
I saw the previous answers of this question on Brainly which were totally rubish, so I reported them and had the moderators remove them,
it's a request to not waste a person's time and effort by giving trash answers just for points.
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