바닷물 1 kg당 마그네슘 1.3 g 존재

바닷물 1 kg당 마그네슘 1.3 g 존재

 

 

마그네슘은 합금, 배터리, 화학 합성에 쓰이는 귀중한 금속이다.

이것은 주로 바닷물에서 얻는데, 바닷물 1 kg당 마그네슘 1.3 g이 존재한다.

마그네슘 8.0×10^4 톤을 얻기 위해서는 바닷물이 얼마나 필요한가?

(바닷물의 밀도는 1.03 g/mL.)

 

 

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in chemical synthesis. It is obtained mostly from seawater, which contains about 1.3 g of Mg for every kilogram of seawater. Calculate the volume of seawater (in liters) needed to extract 8.0×10^4 tons of Mg, which is roughly the annual production in the United States. (Density of seawater = 1.03 g/mL.)

 

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미터법(미터톤)

> 1 ton = 1000 kg

 

 

(8.0×10^4 ton) × (1000 kg / 1 ton) × (1000 g / 1 kg)

= 8.0×10^10 g Mg

 

 

seawater : Mg = 1000 g : 1.3 g = ? g : 8.0×10^10 g

? = 1000 × (8.0×10^10) / 1.3 = 6.153846×10^13 g seawater

 

 

(6.153846×10^13 g) × (1 mL / 1.03 g) × (1 L / 1000 mL)

= 6.0×10^10 L seawater

 

 

답: 6.0×10^10 L

 

 

 

[참고]

미국식(US)으로 계산하면,

> 1 ton = 2000 lb

> 1 lb = 453.6 g

 

(8.0×10^4 ton) × (2000 lb / 1 ton) × (453.6 g / 1 lb)

= 7.2576×10^10 g Mg

 

seawater : Mg = 1000 g : 1.3 g = ? g : 7.2576×10^10 g

? = 1000 × (7.2576×10^10) / 1.3 = 5.582769×10^13 g seawater

 

(5.582769×10^13 g) × (1 mL / 1.03 g) × (1 L / 1000 mL)

= 5.4×10^10 L seawater