Question

The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a concentration of salt within the water, the density of water is 1.03 g/mL. Calculate the total mass of sodium chloride, in grams, contained in all the seawater in the world.

Answer 1

Answer:

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater ($m_{sw}$), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

$m_{sw} = \rho_{sw}\cdot V_{sw}$

Where:

$\rho_{sw}$ - Density of seawater, measured in grams per liters.

$V_{sw}$ - Volume of seawater, measured in liters.

If $V_{sw} = 1.5\times 10^{21}\,L$ and $\rho_{sw} = 1030\,\frac{g}{L}$, then:

$m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)$

$m_{sw} = 1.545\times 10^{24}\,g$

The total mass of sodium chloride is determined by the following ratio:

$r = \frac{m_{NaCl}}{m_{sw}}$

$m_{NaCl} = r\cdot m_{sw}$

Given that $m_{sw} = 1.545\times 10^{24}\,g$ and $r = 0.035$, the total mass of sodium chloride in all the seawater in the world is:

$m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)$

$m_{NaCl} = 5.408\times 10^{22}\,g$

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

Answer 2

Following are the calculation to the total mass of sodium chloride in grams:

Given:

Seawater volume$= 1.5 \times 10^{21}\ L$

Seawater density $= 1.03\ \frac{g}{mL}$

$\%$ of sodium chloride in seawater $= 3.5\ \%$

To find:

The total mass of sodium chloride in grams=?

Solution:

The total volume of seawater $= 1.5 \times 10^{21}\ L= 1.5 \times 10^{24}\ mL$

The density of seawater $= 1.03 \frac{g}{mL}$

$\therefore$

Seawater mass = Seawater volume $\times$ density

                           $= 1.5 \times 10^{24}\times 1.03 \ g\\\\= 1.545 \times 10^{24}\ g$

$\%$ of sodium chloride in seawater $= 3.5\ \%$

$\therefore$

Calculating the sodium chloride mass in Seawater:

$= 3.5\% \ \ of\ \ 1.545 \times 10^{24} \ g\\\\ =\frac{3.5}{100} \times 1.545 \times 10^{24}\ g\\\\= 5.4075 \times 10^{22} \ g$

Calculating the sodium chloride mass $= 5.4075 \times 10^{22} \ g$

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