The density of a pure substance ‘A’, whose atoms are arranged in a cubic close pack arrangement, is $1 g/cc$ . If all the tetrahedral voids are now filled with ‘B’ atoms, what is the density of the resulting solid in $g/cc$ ? [Molar mass $A = 30 g/mol$ and Molar mass $B = 50 g/mol]$

3.33

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4.33

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2.33

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5.33

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Solution

The correct option is B 4.33
We know that density can be given by:
$ρ = \frac{n M}{V N_{A}}$
For pure A, $1 = \frac{4 \times 30}{V N_{A}}$
$⟹ V = \frac{120}{N_{A}}$
$ρ = \frac{4 \times 30 + 8 \times 50}{120}$
$⟹ ρ = 4.33 g / c c$

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The density of a pure substance ‘A’, whose atoms are arranged in a cubic close pack arrangement, is 1g/cc. If all the tetrahedral voids are now filled with ‘B’ atoms, what is the density of the resulting solid in g/cc? [Molar mass A=30g/mol and Molar mass B = 50 g/mol]

The correct option is B 4.33We know that density can be given by: ρ=nMVNA For pure A, 1=4×30VNA ⟹V=120NA ρ=4×30+8×50120 ⟹ρ=4.33g/cc

The density of a pure substance ‘A’, whose atoms are arranged in a cubic close pack arrangement, is $1 g/cc$ . If all the tetrahedral voids are now filled with ‘B’ atoms, what is the density of the resulting solid in $g/cc$ ? [Molar mass $A = 30 g/mol$ and Molar mass $B = 50 g/mol]$

The correct option is B 4.33
We know that density can be given by:
$ρ = \frac{n M}{V N_{A}}$
For pure A, $1 = \frac{4 \times 30}{V N_{A}}$
$⟹ V = \frac{120}{N_{A}}$
$ρ = \frac{4 \times 30 + 8 \times 50}{120}$
$⟹ ρ = 4.33 g / c c$