$\begin{array}{1 1} 37.35\; g\;mol^{-1} \\ 56.02\; g\;mol^{-1} \\ 65.36\; g\;mol^{-1} \\ 74.7\; g\;mol^{-1}\end{array}$

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Answer: $74.7\;g\;mol^{-1}$

The density is given by the formula $\large\frac{z\;M}{a^3\;N_A}$

Given density $\rho=1.984 \;g\;cm^{-3}$ and $a= 630\; ppm$, we need to calculate Molar mass $M$.

Avagadro's number $N_A = 6.022 \times 10^{23} /\; mol$

Since its a face centered cubic cell, $z=4$.

$\Rightarrow M = \large\frac{\rho a^3 N_A}{z} $$=\large\frac{1.984 \times (630\times10^{-10})^3 \times 6.022 \times 10^{23}}{4}$$ = 74.7\;g\;mol^{-1}$

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