Chapter 11, Problem 11.64QP

Interpretation Introduction

Interpretation: The table representing the different ions in sea water is given. The molality of each ion is to be calculated.

Concept introduction: Molality is defined as the number of moles of solute that is present in $1\text{kg}$ of solvent

To determine: The molality of each given ion is to be calculated.

Answer to Problem 11.64QP

Solution

The molarity for given ions in terms of molality is tabulated as,

Ions	Molality $\left(\text{mol/kg}\right)$
${\text{Na}}^{+}$	$0.4753$
${\text{K}}^{+}$	$1.02\times {10}^{-2}$
${\text{Mg}}^{2+}$	$5.3\times {10}^{-2}$
${\text{Ca}}^{2+}$	$1.03\times {10}^{-2}$
${\text{Sr}}^{2+}$	$9.08\times {10}^{-5}$
${\text{Cl}}^{-}$	$5.5\times {10}^{-1}$
${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$	$2.83\times {10}^{-2}$
${\text{HCO}}_3{}^{-}$	$2.06\times {10}^{-3}$
${\text{Br}}^{-}$	$8.5\times {10}^{-4}$
$\text{B}{\left(\text{OH}\right)}_3$	$4.2\times {10}^{-4}$
${\text{F}}^{-}$	$6.9\times {10}^{-5}$

Explanation of Solution

Explanation

Given

The density for sea water is $1.022\text{g}/\text{mL}$

The table for molarity is given as,

Ions	Concentration $\left(\text{mmol}/\text{L}\right)$
${\text{Na}}^{+}$	$480.57$
${\text{K}}^{+}$	$10.46$
${\text{Mg}}^{2+}$	$54.14$
${\text{Ca}}^{2+}$	$10.53$
${\text{Sr}}^{2+}$	$0.0928$
${\text{Cl}}^{-}$	$559.40$
${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$	$28.93$
${\text{HCO}}_3{}^{-}$	$2.11$
${\text{Br}}^{-}$	$0.865$
$\text{B}{\left(\text{OH}\right)}_3$	$0.426$
${\text{F}}^{-}$	$0.070$

Table 1

The conversion of $\text{mmol}$ to $\text{mol}$ is done as,

$1\text{mmol}=0.001\text{mol}$

Hence, the conversion of $480.57\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}480.57\text{mmol}=480.57\times 0.001\text{mol}\\ =480.57\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $10.46\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}10.46\text{mmol}=10.46\times 0.001\text{mol}\\ =10.46\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $54.14\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}54.14\text{mmol}=54.14\times 0.001\text{mol}\\ =54.14\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $10.53\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}10.53\text{mmol}=10.53\times 0.001\text{mol}\\ =10.53\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $0.0928\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}0.0928\text{mmol}=0.0928\times 0.001\text{mol}\\ =0.0928\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $559.40\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}559.40\text{mmol}=559.40\times 0.001\text{mol}\\ =559.40\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $28.93\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}28.93\text{mmol}=28.93\times 0.001\text{mol}\\ =28.93\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $2.11\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}2.11\text{mmol}=2.11\times 0.001\text{mol}\\ =2.11\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $0.865\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}0.865\text{mmol}=0.865\times 0.001\text{mol}\\ =0.865\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $0.426\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}0.426\text{mmol}=0.426\times 0.001\text{mol}\\ =0.426\times {10}^{-3}\text{mol}\end{array}$

Similarly, the conversion of $0.070\text{mmol}$ to $\text{mol}$ is done as,

$\begin{array}{c}0.070\text{mmol}=0.070\times 0.001\text{mol}\\ =0.070\times {10}^{-3}\text{mol}\end{array}$

The concentration of ions is tabulated as,

Ions	Concentration $\left(\text{mol}/\text{L}\right)$
${\text{Na}}^{+}$	$480.57\times {10}^{-3}$
${\text{K}}^{+}$	$10.46\times {10}^{-3}$
${\text{Mg}}^{2+}$	$54.14\times {10}^{-3}$
${\text{Ca}}^{2+}$	$10.53\times {10}^{-3}$
${\text{Sr}}^{2+}$	$0.0928\times {10}^{-3}$
${\text{Cl}}^{-}$	$559.40\times {10}^{-3}$
${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$	$28.93\times {10}^{-3}$
${\text{HCO}}_3{}^{-}$	$2.11\times {10}^{-3}$
${\text{Br}}^{-}$	$0.865\times {10}^{-3}$
$\text{B}{\left(\text{OH}\right)}_3$	$0.426\times {10}^{-3}$
${\text{F}}^{-}$	$0.070\times {10}^{-3}$

Table 2

The molar mass of ${\text{Na}}^{+}$ is $22.99\text{g}/\text{mol}$.

The molar mass of ${\text{K}}^{+}$ is $39.0983\text{g}/\text{mol}$.

The molar mass of ${\text{Mg}}^{2+}$ is $24.30\text{g}/\text{mol}$.

The molar mass of ${\text{Ca}}^{2+}$ is $40.078\text{g}/\text{mol}$.

The molar mass of ${\text{Sr}}^{2+}$ is $87.62\text{g}/\text{mol}$.

The molar mass of ${\text{Cl}}^{-}$ is $35.453\text{g}/\text{mol}$.

The molar mass of ${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$ is $96.06\text{g}/\text{mol}$.

The molar mass of ${\text{HCO}}_3{}^{-}$ is $61.0168\text{g}/\text{mol}$.

The molar mass of ${\text{Br}}^{-}$ is $79.904\text{g}/\text{mol}$.

The molar mass of $\text{B}{\left(\text{OH}\right)}_3$ is $61.831\text{g}/\text{mol}$.

The molar mass of ${\text{F}}^{-}$ is $19.0\text{g}/\text{mol}$.

Number of moles of a substance is given as,

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ (1)

Molarity of a solution is given as,

$\text{Molarity}=\frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$ (2)

Substitute number of moles by mass in equation (2),

$\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}$ (3)

Substitute molar mass of ${\text{Na}}^{+}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 480.57\times {10}^{-3}=\frac{\frac{\text{Given mass}}{22.99\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=11.048\text{g}\end{array}$

Substitute molar mass of ${\text{K}}^{+}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 10.46\times {10}^{-3}=\frac{\frac{\text{Given mass}}{39.0983\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=0.4090\text{g}\end{array}$

Substitute molar mass of ${\text{Mg}}^{2+}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 54.14\times {10}^{-3}=\frac{\frac{\text{Given mass}}{24.30\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=1.316\text{g}\end{array}$

Substitute molar mass of ${\text{Ca}}^{2+}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 10.53\times {10}^{-3}=\frac{\frac{\text{Given mass}}{40.078\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=0.4220\text{g}\end{array}$

Substitute molar mass of ${\text{Sr}}^{2+}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 0.0928\times {10}^{-3}=\frac{\frac{\text{Given mass}}{87.62\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=0.0081\text{g}\end{array}$

Substitute molar mass of ${\text{Cl}}^{-}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 559.40\times {10}^{-3}=\frac{\frac{\text{Given mass}}{35.453\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=19.832\text{g}\end{array}$

Substitute molar mass of ${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 28.93\times {10}^{-3}=\frac{\frac{\text{Given mass}}{96.06\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=2.779\text{g}\end{array}$

Substitute molar mass of ${\text{HCO}}_3{}^{-}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 2.11\times {10}^{-3}=\frac{\frac{\text{Given mass}}{61.0168\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=0.129\text{g}\end{array}$

Substitute molar mass of ${\text{Br}}^{-}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 0.865\times {10}^{-3}=\frac{\frac{\text{Given mass}}{79.904\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=0.0691\text{g}\end{array}$

Substitute molar mass of $\text{B}{\left(\text{OH}\right)}_3$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 0.426\times {10}^{-3}=\frac{\frac{\text{Given mass}}{61.831\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=0.0263\text{g}\end{array}$

Substitute molar mass of ${\text{F}}^{-}$, molarity and volume of solution in equation (3) as,

$\begin{array}{c}\text{Molarity}=\frac{\frac{\text{Given mass}}{\text{Molar mass}}}{\text{Volume of solution in litres}}\\ 0.070\times {10}^{-3}=\frac{\frac{\text{Given mass}}{19.0\frac{\text{g}}{\text{mol}}}}{\text{1}\text{.0}\text{L}}\\ \text{Given mass}=0.0013\text{g}\end{array}$

The mass of all the ions is tabulated as,

Ions	Mass of solute $\left(\text{g}\right)$
${\text{Na}}^{+}$	$11.048$
${\text{K}}^{+}$	$0.4090$
${\text{Mg}}^{2+}$	$1.316$
${\text{Ca}}^{2+}$	$0.4220$
${\text{Sr}}^{2+}$	$0.0081$
${\text{Cl}}^{-}$	$19.832$
${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$	$2.779$
${\text{HCO}}_3{}^{-}$	$0.129$
${\text{Br}}^{-}$	$0.0691$
$\text{B}{\left(\text{OH}\right)}_3$	$0.0263$
${\text{F}}^{-}$	$0.0013$

Table 3

The conversion of $\text{g}$ to $\text{kg}$ is done as,

$1\text{g}=0.001\text{kg}$

Hence, the conversion of $1.022\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}1.022\text{g}=1.022\times 0.001\text{kg}\\ =1.022\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $11.048\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}11.048\text{g}=11.048\times 0.001\text{kg}\\ =11.048\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $0.4090\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}0.4090\text{g}=0.4090\times 0.001\text{kg}\\ =0.4090\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $1.316\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}1.316\text{g}=1.316\times 0.001\text{kg}\\ =1.316\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $0.4220\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}0.4220\text{g}=0.4220\times 0.001\text{kg}\\ =0.4220\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $0.0081\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}0.0081\text{g}=0.0081\times 0.001\text{kg}\\ =0.0081\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $19.832\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}19.832\text{g}=19.832\times 0.001\text{kg}\\ =19.832\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $2.779\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}2.779\text{g}=2.779\times 0.001\text{kg}\\ =2.779\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $0.129\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}0.129\text{g}=0.129\times 0.001\text{kg}\\ =0.129\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $0.0691\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}0.0691\text{g}=0.0691\times 0.001\text{kg}\\ =0.0691\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $0.0263\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}0.0263\text{g}=0.0263\times 0.001\text{kg}\\ =0.0263\times {10}^{-3}\text{kg}\end{array}$

Similarly, the conversion of $0.0013\text{g}$ to $\text{kg}$ is done as,

$\begin{array}{c}0.0013\text{g}=0.0013\times 0.001\text{kg}\\ =0.0013\times {10}^{-3}\text{kg}\end{array}$

The conversion of $\text{mL}$ to $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

The mass of $0.001\text{}\text{L}$ solution $=1.022\times {10}^{-3}\text{kg}$

The mass of $1\text{}\text{L}$ solution $\begin{array}{c}=\frac{1.022\times {10}^{-3}}{0.001}\text{kg}\\ \text{=1}\text{.022}\text{kg}\end{array}$

Mass of solvent is given as,

$\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}$ (4)

Substitute the mass of solution and mass of ${\text{Na}}^{+}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-11.048\times {10}^{-3}\text{kg}\\ \text{=1}\text{.011}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{K}}^{+}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-0.4090\times {10}^{-3}\text{kg}\\ \text{=1}\text{.022}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{Mg}}^{2+}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-1.316\times {10}^{-3}\text{kg}\\ \text{=1}\text{.021}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{Ca}}^{2+}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-0.4220\times {10}^{-3}\text{kg}\\ \text{=1}\text{.022}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{Sr}}^{2+}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-0.0081\times {10}^{-3}\text{kg}\\ \text{=1}\text{.022}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{Cl}}^{-}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-19.832\times {10}^{-3}\text{kg}\\ \text{=1}\text{.002}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-2.779\times {10}^{-3}\text{kg}\\ \text{=1}\text{.02}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{HCO}}_3{}^{-}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-0.129\times {10}^{-3}\text{kg}\\ \text{=1}\text{.02}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{Br}}^{-}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-0.0691\times {10}^{-3}\text{kg}\\ \text{=1}\text{.02}\text{kg}\end{array}$

Substitute the mass of solution and mass of $\text{B}{\left(\text{OH}\right)}_3$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-0.0263\times {10}^{-3}\text{kg}\\ \text{=1}\text{.02}\text{kg}\end{array}$

Substitute the mass of solution and mass of ${\text{F}}^{-}$ solute in equation (4),

$\begin{array}{c}\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}\\ \text{=1022}\times {10}^{-3}\text{kg}-0.0013\times {10}^{-3}\text{kg}\\ \text{=1}\text{.02}\text{kg}\end{array}$

The mass of solvent for the given ions that act as solute is tabulated as,

Ions	Mass of solvent $\left(\text{kg}\right)$
${\text{Na}}^{+}$	$\text{1}\text{.011}$
${\text{K}}^{+}$	$\text{1}\text{.022}$
${\text{Mg}}^{2+}$	$\text{1}\text{.021}$
${\text{Ca}}^{2+}$	$\text{1}\text{.022}$
${\text{Sr}}^{2+}$	$\text{1}\text{.022}$
${\text{Cl}}^{-}$	$\text{1}\text{.002}$
${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$	$\text{1}\text{.02}$
${\text{HCO}}_3{}^{-}$	$\text{1}\text{.02}$
${\text{Br}}^{-}$	$\text{1}\text{.02}$
$\text{B}{\left(\text{OH}\right)}_3$	$\text{1}\text{.02}$
${\text{F}}^{-}$	$\text{1}\text{.02}$

Table 4

The molality of the solution is given as,

$\text{Molality}=\frac{\text{Numbe of moles of solute}}{\text{Mass of solvent in kg}}$ (5)

Substitute the value of number of moles of ${\text{Na}}^{+}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{\text{480}\text{.57}\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.011}\text{kg}}\\ =\underset{\_}{0.4753mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{K}}^{+}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{\text{10}\text{.46}\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{1.02\times 10^{-2}mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{Mg}}^{2+}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{\text{54}\text{.14}\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.021}\text{kg}}\\ =\underset{\_}{5.3\times 10^{-2}mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{Ca}}^{2+}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{\text{10}\text{.53}\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{1.03\times 10^{-2}mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{Sr}}^{2+}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{0.0928\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{9.08\times 10^{-5}mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{Cl}}^{-}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{559.40\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{5.5\times 10^{-1}mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{28.93\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{2.83\times 10^{-2}mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{HCO}}_3{}^{-}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{2.11\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{2.06\times 10^{-3}mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{Br}}^{-}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{0.865\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{8.5\times 10^{-4}mol/kg}\end{array}$

Substitute the value of number of moles of $\text{B}{\left(\text{OH}\right)}_3$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{0.426\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{4.2\times 10^{-4}mol/kg}\end{array}$

Substitute the value of number of moles of ${\text{F}}^{-}$ and mass of solvent in equation (5).

$\begin{array}{c}\text{Molality}=\frac{0.070\times {\text{10}}^{-}{}^3\text{mol}}{\text{1}\text{.022}\text{kg}}\\ =\underset{\_}{6.9\times 10^{-5}mol/kg}\end{array}$

The molality of the given ions is tabulated as,

Ions	Molality $\left(\text{mol/kg}\right)$
${\text{Na}}^{+}$	$0.4753$
${\text{K}}^{+}$	$1.02\times {10}^{-2}$
${\text{Mg}}^{2+}$	$5.3\times {10}^{-2}$
${\text{Ca}}^{2+}$	$1.03\times {10}^{-2}$
${\text{Sr}}^{2+}$	$9.08\times {10}^{-5}$
${\text{Cl}}^{-}$	$5.5\times {10}^{-1}$
${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$	$2.83\times {10}^{-2}$
${\text{HCO}}_3{}^{-}$	$2.06\times {10}^{-3}$
${\text{Br}}^{-}$	$8.5\times {10}^{-4}$
$\text{B}{\left(\text{OH}\right)}_3$	$4.2\times {10}^{-4}$
${\text{F}}^{-}$	$6.9\times {10}^{-5}$

Table 4

Conclusion

The molality for the given ions of sea water is tabulated as,

Ions	Molality $\left(\text{mol/kg}\right)$
${\text{Na}}^{+}$	$0.4753$
${\text{K}}^{+}$	$1.02\times {10}^{-2}$
${\text{Mg}}^{2+}$	$5.3\times {10}^{-2}$
${\text{Ca}}^{2+}$	$1.03\times {10}^{-2}$
${\text{Sr}}^{2+}$	$9.08\times {10}^{-5}$
${\text{Cl}}^{-}$	$5.5\times {10}^{-1}$
${\text{SO}}_{\text{4}}{{}^{\text{2}}}^{-}$	$2.83\times {10}^{-2}$
${\text{HCO}}_3{}^{-}$	$2.06\times {10}^{-3}$
${\text{Br}}^{-}$	$8.5\times {10}^{-4}$
$\text{B}{\left(\text{OH}\right)}_3$	$4.2\times {10}^{-4}$
${\text{F}}^{-}$	$6.9\times {10}^{-5}$

Table 4