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|5. Gold metal crystallizes in a face -centred cubicunit 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 গঠনকৰে। গ’ল্ডৰ ঘনত্ব নির্ণয় কৰা। (গ’ল্ডৰ পাৰমাণৱিক ভৰ |=179 u, পাৰমাণৱিক ব্যাসার্ধ =0.144nm, |
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Answer» Answer: ✨ ✨༺ H E Y 彡M A T E ࿐✨ ✨ Here Is Your Answer, Hope It helps ___________________________________________ 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 ---------------------------------------------------------------------------------- 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. ___________________________________________ , (づ￣ ³￣)づ |
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