A 4.10 lb sample of the mineral luberoite has 65.3% gold. How many meters of gold wire with diameter of 0.0620 in can be produced from this sample? Density of gold 19.3 g/cm3.

Question

Recall: V(cylinder) = π*r²*h and 1lb=454g.

Answer Choices:

8.13 m

82.1 m

209. m

32.5 m (Correct answer)

3250 m

Answer 1

Find mass of gold in grams:

4.10 lb x 65.3% = 2.677 lbs

2.677 lbs x 454 g / lb = 1215 g

Find the volume of gold:

1215 g x 1 cm³ / 19.3 g = 62.95 cm³

Find the length (height) of wire:

r = 1/2 d = 1/2 x 0.0620 inch = 0.0310 inch x 2.54 cm / inch = 0.07874 cm

V = π r² h

62.95 cm³ = (3.1416)(0.07874 cm)² h

h = 3232 cm

Converting this to meters, we have..

h = 3232 cm x 1 m / 100 cm = 32.3 m (differs in the tenths place probably due to conversion factor or rounding)