An element with molar mass 2.7×10⁻² kg mol⁻¹ forms a cubic unit cell with edge length 405 pm. If its density i

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2.7\times10^{-2}

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Step 1 – Density equation.

d=\frac{z\,M}{a^3\,N_A}\quad\Rightarrow\quad z=\frac{d\,a^3\,N_A}{M}.

Step 2 – Put quantities in consistent (CGS) units.

a=405\text{ pm}=4.05\times10^{-8}\text{ cm},\quad a^3=6.643\times10^{-23}\text{ cm}^3,

M=2.7\times10^{-2}\text{ kg mol}^{-1}=27\text{ g mol}^{-1},\quad d=2.7\times10^{3}\text{ kg m}^{-3}=2.7\text{ g cm}^{-3}.

Step 3 – Substitute.

z=\frac{(2.7\text{ g cm}^{-3})(6.643\times10^{-23}\text{ cm}^3)(6.022\times10^{23}\text{ mol}^{-1})}{27\text{ g mol}^{-1}}.

Numerator $=2.7\times6.643\times6.022=108.0$ (the $10^{-23}$ and $10^{23}$ cancel).

$z=\frac{108.0}{27}=4.0.$ …

Q.An element with molar mass $2.7\times10^{-2}$ kg mol $^{-1}$ forms a cubic unit cell with edge length 405 pm. If its density is $2.7\times10^{3}$ kg m $^{-3}$ , what is the nature of the cubic unit cell?