Answer
m = 105.5 g is the mass of helium in the weather balloon.
Work Step by Step
First we have to find the volume of the sphere of helium gas and after that we find the mass of helium.
$V_{sphere}$ = $\frac{4}{3}$ π $r^{3}$
The diameter of the helium filled weather balloon is 3.50 ft, so its radius is $\frac{3.50}{2}$ ft = 1.75 ft.
Converted to inches this radius is: 1.75 ft x $\frac{12 in}{1 ft}$ = 21 in
The radius converted to cm will be:
21 in = 21 in x $\frac{2.54 (cm)}{1 in}$
21 in = 53.34 cm
$V_{sphere}$ = $\frac{4}{3}$x 3.14 x$(53.34cm)^{3}$
$V_{sphere}$ = 635,371 $cm^{3}$
$V_{sphere}$ = 635,371$cm^{3}$ x $\frac{1 L}{(10)^{3}cm^{3}}$
$V_{sphere}$ = 635,371 x ${10^{-3}L}$
Now we find the mass: m = d x V
m = 0.166 $\frac{g}{L}$ x 635,371 x ${10^{-3}L}$
m = 105,471 x ${10^{-3}}$ g
or m = 105.5 g
