Respuesta :
Given that, 1.30 g of mg per kg (1000 g) of sea water.
Convert [tex] 5.33\times 10^{4}tons [/tex] into gram.
1 ton = 2000 lb
Thus,
[tex] 5.33\times 10^{4}tons [/tex] = [tex] 2000 lb \times5.33\times 10^{4}tons [/tex]
= [tex] 10660 \times 10^{4} lb [/tex] = [tex] 1.0660 \times 10^{8} lb [/tex]
1 lb = 453.6 g
Thus,
[tex] 1.0660 \times 10^{8} lb [/tex] = [tex] 1.0660 \times 10^{8}\times 453.6 g [/tex]
= [tex] 483.5376 \times 10^{8} g [/tex] = [tex] 4.83 \times 10^{10} g [/tex]
Now,
1.30 g of magnesium is present in 1000 g of seawater.
Thus, 1 g of magnesium present in [tex] \frac{1000 g}{1.30} [/tex] of seawater.
[tex] 4.83 \times 10^{10} g [/tex] of magnesium present in [tex] 4.83 \times 10^{10} g\times \frac{1000 g}{1.30} [/tex] = [tex] 3715.38 \times 10^{10} g [/tex]
= [tex] 3.715 \times 10^{13} g [/tex] of seawater.
Now, density of sea water is [tex] 1.03 g/ml [/tex]
Density is equal to the ratio of mass and volume.
[tex] Density =\frac{Mass}{Volume} [/tex]
Put the value of density and mass to find the volume:
[tex] 1.03 g/ml =\frac{3.715 \times 10^{13} g}{Volume} [/tex]
Volume = [tex]\frac{3.715 \times 10^{13} g}{1.03 g/ml} [/tex]
Volume = [tex] 3.606\times 10^{13}ml [/tex]
Now, convert ml into L:
1 L = 1000 ml
Volume in L = [tex] \frac{3.606\times 10^{13}}{1000 ml} [/tex]
= [tex] 3.606\times 10^{10}L [/tex]
Thus, volume in litres = [tex] 3.606\times 10^{10}L [/tex]
