Question

Magnesium occurs in seawater to the extent of 1.4 g magnesium per kilogram of seawater. What volume of seawater, in cubic meters, would have to be processed to produce 1.00 x 105 tons of magnesium (1 ton = 2000 lb)? Assume a density of 1.025 g/ml for seawater.

Answer 1

Answer: $6.3\times 10^7m^3$

Explanation:

Required amount of magnesium = $1.00\times 10^5 tons$

Given : 1 ton = 2000 lb

$1.00\times 10^5 tons=\frac{2000}{1}\times 1.00\times 10^5=2\times 10^8lb$

1 lb = 453.592 g

$2\times 10^8 lb=\frac{453.592 }{1}\times 2\times 10^8 lb=907.184\times 10^8g$

Given : 1.4 g of magnesium is produced by 1000 g of sea water

$907.184\times 10^8g$ of magnesium is produced by =$\frac{1000}{1.4}\times 907.184\times 10^8g=6.5\times 10^{13}$ g of sea water

Density of sea water  = 1.025 g/ml

Volume of sea water =$\frac{\text {mass of sea water}}{\text {density of sea water}}=\frac{6.5\times 10^{11}g}{ 1.025g/ml}=6.3\times 10^{13}ml$

1 ml = $10^{-6}m^3$

$6.3\times 10^{13}ml=\frac{10^{-6}}{1}\times 6.3\times 10^{13}=6.3\times 10^7m^3$

Volume of seawater, in cubic meters is $6.3\times 10^7$