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Question: How many grams of NaCl are contained in 350 ml of a 0.287 M solution of sodium chloride?...

How many grams of NaCl are contained in 350 ml of a 0.287 M solution of sodium chloride?

Explanation

Solution

This question can be solved using the mole concept and the concept of molarity. The molarity of a solution is defined as the concentration of a solute in a solution is in terms of the amount of solute per unit volume, or the molar concentration.

Complete answer:
Molarity can be given by
M=wsMs×1000VM=\dfrac{{{w}_{s}}}{{{M}_{s}}}\times \dfrac{1000}{V}
Where ws{{w}_{s}}= mass of the solute dissolved (in grams),
Ms{{M}_{s}}= molar mass of the solute dissolved (in grams),
And V= total volume of the solution (in mL).
The SI unit for molarity is mol/m3{{m}^{3}} but it is usually expressed in mol/dm3d{{m}^{3}} which is equivalent to mol/L. It can also be expressed by the capital letter M.
To calculate the mass of NaCl present in the solution, we first need to determine the molecular weight of NaCl.
Now, the molecular mass of one mole of a compound is equal to the sum of atomic masses of all the elements present in the molecule.
We know that the atomic mass of sodium (MNa{{M}_{Na}}) is 22.99 g/mol and the atomic mass of chlorine (MCl{{M}_{Cl}}) is 35.45 g/mol
So, the molecular mass or the mass of one mole of NaCl (MNaCl{{M}_{NaCl}}) is

& {{M}_{NaCl}}={{M}_{Na}}+{{M}_{Cl}} \\\ & {{M}_{NaCl}}=22.99+35.45 \\\ & {{M}_{NaCl}}=58.44g/mol \\\ \end{aligned}$$ It is given to us that the molarity of the sodium chloride solution M=0.287 mol/L and the volume of the solution is V=350 mL. Upon substituting these values in the molarity equation, we get $$\begin{aligned} & M=\dfrac{{{w}_{NaCl}}}{{{M}_{NaCl}}}\times \dfrac{1000}{V} \\\ & {{w}_{NaCl}}=\dfrac{M\times {{M}_{NaCl}}\times V}{1000} \\\ & {{w}_{NaCl}}=\dfrac{0.287\times 58.44\times 350}{1000} \\\ & {{w}_{NaCl}}\cong 5.87g \\\ \end{aligned}$$ So, approximately 5.87 grams of NaCl are contained in 350 ml of a 0.287 M solution of sodium chloride. **Note:** It should be noted that if one mole of a substance is present, it has exactly the Avogadro number (${{N}_{A}}=6.022\times {{10}^{23}}$) of particles and the mass of one mole of a compound is equivalent to the sum of the mass of all the particles contained in one mole of a substance. This can also be used to define the molar mass of a substance.