Question

Most abundant element dissolved in seawater in chlorine at a concentration of $19g/kg$ of sea water. The volume of earth’s ocean is $1.4 \times {10^{21}}l$. How many g-atoms of chlorine are potentially available from the oceans? Density of seawater is $1g/ml$. $\left( {{N_A} = 6 \times {{10}^{23}}} \right)$
A. $7.5 \times {10^{20}}$
B. $27 \times {10^{21}}$
C. $27 \times {10^{24}}$
D. $7.5 \times {10^{19}}$

Answer 1

We have to first calculate the total mass of the ocean using the volume of earth’s ocean and density of seawater. Then we have to calculate the total weight of the chloride ions and from the calculated total weight of the chloride ions we have to calculate the number of gram atoms of chlorine.

Complete step by step solution:
Given data contains,
Concentration of chlorine in seawater is $19g/kg$.
Volume of the earth’s ocean is $1.4 \times {10^{21}}l$.
Density of seawater is $1g/ml$.
Let us now calculate the total weight of the ocean. We have to multiply the volume of earth’s ocean and density of seawater to get the total weight of the ocean.
Total weight of the ocean=${\text{Volume of earth ocean}} \times {\text{Density of seawater}}$
We can now substitute the values of volume of earth’s ocean and the density of seawater in the formula that is used to calculate the total weight of the ocean.
Total weight of the ocean=$1.4 \times {10^{21}}\not{L} \times 1 \times {10^3}\,\dfrac{g}{{\not{L}}}$
Total weight of the ocean=$1.4 \times {10^{24}}g$
The total weight of the ocean is $1.4 \times {10^{24}}g$.
Let us now calculate the concentration of chlorine in water.
$\left[ {C{l^ - }} \right] = \dfrac{{19gC{l^ - }}}{{1000g{\text{water}}}}$
$\left[ {C{l^ - }} \right] = 19 \times {10^{ - 3}}gCl/gwater$
The concentration of chlorine in water is $19 \times {10^{ - 3}}gCl/gwater$.
We can calculate the total weight of chloride by taking the product of total weight of ocean and concentration of chlorine in water.
Total weight of chloride=$1.4 \times {10^{24}}\not{{gwater}} \times 19 \times {10^{ - 3}}\dfrac{{gC{l^ - }}}{{\not{{gwater}}}}$
Total weight of chloride=$2.66 \times {10^{22}}gC{l^ - }$
So, the total weight of chloride is $2.66 \times {10^{22}}gC{l^ - }$.
From the total weight of chloride, we have to calculate the total number of grams of atoms using the mass of chlorine.
We know that the mass of chlorine atoms is $35.45$.
So, the number of gram atoms of chlorine is $\dfrac{{2.66 \times {{10}^{22}}}}{{35.45}}$
The number of gram atoms of chlorine=$7.5 \times {10^{20}}g$atoms.
The number of gram atoms of chlorine is $7.5 \times {10^{20}}g$atoms.
Therefore,option (A) is correct.

Note:
An alternate approach to solve this problem is,
Given data contains,
Concentration of chlorine in seawater is $19g/kg$.
Volume of the earth’s ocean is $1.4 \times {10^{21}}l$.
Density of seawater is $1g/ml$.
Let us now calculate the total weight of the ocean. We have to multiply the volume of earth’s ocean and density of seawater to get the total weight of the ocean.
Total weight of the ocean=${\text{Volume of earth ocean}} \times {\text{Density of seawater}}$
We can now substitute the values of volume of earth’s ocean and the density of seawater in the formula that is used to calculate the total weight of the ocean.
Total weight of the ocean=$1.4 \times {10^{21}}\not{L} \times 1 \times {10^3}\,\dfrac{g}{{\not{L}}}$
Total weight of the ocean=$1.4 \times {10^{24}}g$
The total weight of the ocean is $1.4 \times {10^{21}}kg$.
Let us now calculate the moles of chlorine in seawater.
Molar mass of $C{l_2}$ is $71g/mol$.
Moles of chlorine=$\dfrac{{19}}{{71}}$
Moles of chlorine=$0.267mol$
The moles of chlorine is $0.267mol$
The moles of $C{l_2}$ present in $1.4 \times {10^{21}}kg$ is $1.4 \times {10^{21}} \times 0.267 = 0.375 \times {10^{21}}$
The moles of chlorine present in kg of water is $0.375 \times {10^{21}}mol$.
We can calculate the g-atoms of chlorine by multiplying the moles two times.
Gram-atoms of $C{l_2}$=$2 \times 0.375 \times {10^{21}}$
Gram-atoms of $C{l_2}$=$0.75 \times {10^{21}}$
Gram-atoms of $C{l_2}$=$7.5 \times {10^{20}}$
The gram atoms of $C{l_2}$ is $7.5 \times {10^{20}}$
Therefore,option (A) is correct.