Question

Describe how would you prepare 1 L of a 1 M solution of sodium chloride. The gram formula weight of sodium chloride is 58.44 g/mol.

Answer 1

Hint: Here, we have to calculate the mass of Sodium chloride (NaCl) needed to prepare 1 Molar solution of NaCl and the volume of the solution. We have to use the formula of molarity, that is, Molarity=$\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}
{{{\text{Volume}}\,{\text{of}}\,{\text{solution}}}}$.

Complete step by step answer:
First, we have to calculate the moles of NaCl. The molarity of the solution is 1 M and the volume of the solution is 1 L. So, moles of NaCl can be calculated as,
Moles of NaCl=Molarity $ \times $
Volume of solution= 1L
Moles of NaCl= 1 M $ \times $ 1L=1 mol
So, the mole of NaCl in the solution is 1 mol.
Now, we have to calculate the mass of NaCl using the number of moles of NaCl.
Number of moles=$\dfrac{{{\text{Mass}}}}
{{{\text{Molar}}\,{\text{mass}}}}$
The mole of NaCl is 1 mol and the molar mass of NaCl is 58.44 g/mol.
$ \Rightarrow $1=$\dfrac{{{\text{Mass}}}}{{58.44\,{\text{g/mol}}}}$
$ \Rightarrow {\text{Mass}}\,{\text{of}}\,{\text{NaCl}} = 58.44\,{\text{g}}$
So, we need 58.44 g of NaCl to make 1 M solution.
Now, we discuss how to prepare one molar solution of NaCl. A 1M solution of NaCl can be prepared by measuring 58.44 g of NaCl and placing this amount of salt in 1 litre volumetric flask and then filling the flask with distilled water to the graduation mark.

Note: It is to be noted that molarity and molality are not the same. Molarity is the moles of solute present in 1 litre of the solution and the molarity is the moles of solute present in 1kg of solvent. Symbolically molarity is represented by M and molality is represented by m.