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=${\displaystyle \frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}}$.

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=${\displaystyle \frac{\text{Mass}}{\text{Molar}\text{mass}}}$
The mole of NaCl is 1 mol and the molar mass of NaCl is 58.44 g/mol.
$\Rightarrow$1=${\displaystyle \frac{\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.