Question

Units of parts per million (ppm) or parts per billion (ppb) are often used to describe the concentrations of solutes in very dilute solutions. The units are defined as the number of grams of solute per million or per billion grams of solvent. Bay of Bengal has $1.89{\text{ppm}}$ of lithium ions. The molality of ${\text{L}}{{\text{i}}^ + }$ in this water is (atomic mass of ${\text{Li}} = 7$):
A.$1.5 \times {10^{ - 4}}{\text{m}}$
B.$1.7 \times {10^{ - 4}}{\text{m}}$
C.$2.5 \times {10^{ - 4}}{\text{m}}$
D.$2.7 \times {10^{ - 4}}{\text{m}}$

Answer 1

Hint: ppm units are the number of grams of a solute present in a million grams of solvent. While the molality is defined as the number of moles of solute present in a thousand grams of solvent.
Formula used
 ${\text{Molality(m)}} = \dfrac{{{\text{moles of lithium ions (n)}}}}{{{\text{mass of solvent(in kg)}}}}$

Complete step by step answer:
We are given the question that Bay of Bengal has $1.89{\text{ppm}}$ of lithium ions. This means $1.89{\text{g}}$ of lithium ions are present in $1000000{\text{g}}$ of the solvent.
So we can say that $1{\text{g}}$ of solvent contains $1.89 \times {10^{ - 6}}{\text{g}}$ of lithium ions.
Thus, the amount of solute present in a thousand grams of solvent (one kilogram) is $1.89 \times {10^{ - 3}}$.
So the number of moles of lithium ions present in one kilogram of solvent is equal to: ${\text{n}} = \dfrac{{1.89 \times {{10}^{ - 3}}}}{7} = 2.7 \times {10^{ - 4}}$moles.
The molality of lithium ions in the water is given by the number of moles of the lithium ions present per kilogram of water.
${\text{Molality(m)}} = \dfrac{{{\text{moles of lithium ions (n)}}}}{{{\text{mass of solvent(in kg)}}}}$
Substituting the values, we get
${\text{m}} = \dfrac{{2.7 \times {{10}^{ - 4}}}}{1}$
$\therefore {\text{m}} = 2.7 \times {10^{ - 4}}{\text{m}}$

Thus, the correct option is D.

Note:
Parts-per-unit notation is generally used for describing dilute solutions, for example, the relative abundance of various dissolved minerals or pollutants present in water. The quantity $1{\text{ppm}}$ may be used as a mass fraction if a pollutant in water is present in the amount of one-millionth of a gram per gram of the water sample solution.
In case of aqueous solutions, we commonly assume the density of water to be $1{\text{g/L}}$. Therefore, we generally equate one kilogram of water with one liter of water. Thus, $1{\text{ppm}}$ corresponds to $1{\text{g/L}}$ and $1{\text{ppb}}$ corresponds to $1\mu {\text{g/L}}$.
All parts per unit notation are dimensionless quantities as they are the ratios of the quantities of the same unit. And, we know that in a mathematical expression, the units of measurement always cancel when in fraction.