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# How many grams of NaCl should be weighed to prepare 1 L of 20 ppm solution of ${\text{N}}{{\text{a}}^ + }$

Last updated date: 09th Sep 2024
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Hint: Parts per million (ppm) is the number of parts by mass (or by volume) of one component per million parts by mass (or by volume) of the solution. For example, if A and B are the two components of the solution. Then,
${\text{pp}}{{\text{m}}_{\text{A}}} = \dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{A}}}} {{{\text{Mass}}\,{\text{of}}\,{\text{Solution}}\,}} \times {10^6}$

Here, first we have to calculate the mass of water using the formula of density. The volume of water is given as 1 L and the density of the water is 1g/mL or 1000 g/L
The formula of density is,
Density=$\dfrac{{{\text{Mass}}}} {{{\text{Volume}}}}$
Mass of water=$1000\,\dfrac{{\text{g}}} {{\text{L}}} \times \,{\text{L = 1000}}\,{\text{g}}$
We know that the sodium chloride undergoes complete dissociation to form sodium ions and chloride ions.
${\text{NaCl}}\left( {aq} \right) \to {\text{N}}{{\text{a}}^ + }\left( {aq} \right) + {\text{C}}{{\text{l}}^ - }\left( {aq} \right)$
Here, the mole ratio between sodium chloride and sodium cation is 1:1. This mole ratio will help us to calculate the mass of sodium chloride needed to dissolve in order to get the particular solution.
The concentration of the target solution (in ppp) is,
${\text{ppm = }}\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}} {{{\text{Mass}}\,{\text{of}}\,{\text{water}}}} \times {10^6}$
Mass of solute$= \dfrac{{ppm \times {\text{Mass}}\,{\text{of}}\,{\text{water}}}} {{{{10}^6}}}$
The given value of mass of water is 1 L and the ppm is 20. So, we have to put these values in the above equation.
Mass of solute$= \dfrac{{ppm \times {\text{Mass}}\,{\text{of}}\,{\text{water}}}} {{{{10}^6}}}$
$\Rightarrow$Mass of solute=$\dfrac{{20 \times 1000}} {{{{10}^6}}} = 0.02\,{\text{g}}$
That means the solution contains 0.02 g of sodium cations, ${\text{N}}{{\text{a}}^ + }$. Now, we calculate the moles of sodium cations present in the solution.
Molar mass of sodium is $23\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$.
Moles of sodium cations=$\dfrac{{{\text{Mass}}}} {{{\text{Molar}}\,{\text{mass}}}} = \dfrac{{0.02}} {{23}} = 0.000879\,{\text{mole}}$
The moles ratio of sodium chloride and sodium cation is 1:1. So, the moles of sodium chloride is also 0.000879.
Now, we have to use the formula of the number of moles to calculate the mass of sodium chloride.
Moles of sodium chloride=$\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{sodium}}\,{\text{chloride}}}} {{{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{sodium}}\,{\text{chloride}}}}$
The moles of NaCl is 0.000879
The molar mass of NaCl=$23 + 35.5 = 58.5\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
Now, we have to put all the values in the above equation.
Mass of sodium chloride= Moles of sodium chloride$\times$
Molar mass of sodium chloride
$\Rightarrow$Mass of NaCl=$0.000879 \times 58.5 = 0.051\,{\text{g}}$

Hence, 0.051 grams of NaCl should be weighed to prepare 1 L of 20 ppm solution of ${\text{N}}{{\text{a}}^ + }$.

Note: It is to be noted that for very dilute solutions, terms parts per billion $\left( {{{10}^9}} \right)$ is used to express the concentration. Concentration in parts per thousand (ppt) is also quite often used. Degree of hardness of water is normally expressed as ppm.