Answer
$m=0.933Kg$
Work Step by Step
We know that;
$m=Mn$..............eq(1)
$n=7.50\times 10^{24}atoms\times(\frac{1 mol}{6.023\times 10^{23}atoms})=12.45mol$
We plug in the known values in eq(1) to obtain:
$m=(74.9\frac{g}{mol})(12.45)=932.5g=933g=0.933Kg$
