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CBSE Class 12-science Answered

An element X' (At. mass a 40 g mol-1) having f.c.c. structure, has unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 x 1023 mol-1)
Asked by raul.chintakindi14 | 14 Mar, 2018, 11:44: AM
Expert Answer
Unit space cell space edge space length space equals space 400 space pm

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 400 cross times 10 to the power of negative 10 end exponent space cm

Volume space of space unit space cell space equals space straight a cubed

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open parentheses 400 cross times 10 to the power of negative 10 end exponent space cm close parentheses cubed

Volume space of space unit space cell space space equals space 64 cross times 10 to the power of negative 24 end exponent space cm cubed
space
We space know space that comma

No. space of space unit space cells space in space fcc space equals space 4

Mass space of space 1 space atom space equals fraction numerator Atomic space mass over denominator Avogadro apostrophe straight s space no. end fraction

space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 40 over denominator 6.022 cross times 10 to the power of 23 end fraction

Mass space of space unit space cell space equals space No. space of space atoms space in space unit space cell space cross times space Mass space of space each space atom

space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 space cross times fraction numerator 40 over denominator 6.022 cross times 10 to the power of 23 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space equals 26.57 cross times 10 to the power of negative 23 end exponent space gmol to the power of negative 1 end exponent

Density space of space unit space cell space equals fraction numerator Mass space of space unit space cell space over denominator Volume space of space unit space cell end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 26.57 cross times 10 to the power of negative 23 end exponent over denominator 64 cross times 10 to the power of 24 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 4.15 space gcm to the power of negative 3 end exponent

26.57 space cross times 10 to the power of negative 23 space end exponent gm space cotains space 1 space unit space cell

therefore space 4 space straight g space equals space fraction numerator 1 cross times 4 over denominator 26.57 space cross times 10 to the power of negative 23 space end exponent end fraction

space space space space space space space space space space space space equals space 0.15 space cross times 10 to the power of 23 space end exponent

space Number space of space unit space cells space in space 4 space gm space are space 15 cross times 10 to the power of 21.
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