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* 10^7m^3`

Step-by-step explanation:

Required amount of magnesium =
`1.00* 10^5 tons`

Given : 1 ton = 2000 lb

`1.00* 10^5 tons=(2000)/(1)* 1.00* 10^5=2* 10^8lb`

1 lb = 453.592 g

`2* 10^8 lb=(453.592 )/(1)* 2* 10^8 lb=907.184* 10^8g`

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

`907.184* 10^8g` of magnesium is produced by =
`(1000)/(1.4)* 907.184* 10^8g=6.5* 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}}=(6.5* 10^(11)g)/( 1.025g/ml)=6.3* 10^(13)ml`

1 ml =
`10^(-6)m^3`

`6.3* 10^(13)ml=(10^(-6))/(1)* 6.3* 10^(13)=6.3* 10^7m^3`

Volume of seawater, in cubic meters is
`6.3* 10^7`