Question

How many grams of NaCl in 50 mL of $ {{0}}{{.75}} $ M NaCl solution?

Answer 1

Hint: In the above question, we are asked to find out weight of NaCl present in 50 mL of $ {{0}}{{.75}} $ M NaCl solution. From the molarity, we can find the number of moles of NaCl present in the given solution. After getting the number of moles and calculating molar mass of NaCl, weight of NaCl can be found out.

Formula Used
 $ {{M = }}\dfrac{{{n}}}{{{V}}} $
where M = molarity
n= number of moles of solute
V= volume of solution in liter.

Complete step by step solution
We know that molarity can be defined as the ratio of the number of solutes present in 1 liter of solution. Mathematically, it can be represented as:
 $ {{M = }}\dfrac{{{n}}}{{{V}}} $
Volume is given as 50 mL = $ {{0}}{{.05}} $ L
Rearranging the above equation, we get:
 $ {{n = M \times V}} $
Substituting the values, we get:
 $ {{n = M \times V = 0}}{{.75 \times 0}}{{.05 = 0}}{{.0375}} $
Number of moles of NaCl present is $ {{0}}{{.0375}} $ .
We know that the number of moles can be defined as the ratio of given mass by molar mass. So, let us find out the molar mass of NaCl.
Molar mass of NaCl = atomic mass of Na + atomic mass of Cl = $ {{23 + 35}}{{.5 = 58}}{{.5}} $
So, the weight of NaCl present = number of moles of NaCl $ {{ \times }} $ molar mass of NaCl = $ {{0}}{{.0375 \times 58}}{{.5 = 2}}{{.19}} $ g
Therefore, $ {{2}}{{.19}} $ g of of NaCl is present in 50 mL of $ {{0}}{{.75}} $ M NaCl solution.

Note
Sometimes, there is a confusion between molarity and molality. In such a case, remember, the difference between molarity and molality lies in the denominator part of the ratio. In molarity, we divide the number of moles by volume of the solution. And in molality, we divide the number of moles by mass of solvent in kg.
Solution=Solute + Solvent
So, in molality solute mass must be subtracted to get the correct result.
In molarity, the unit of the volume of the solution should be taken into consideration. It should be converted into liters in order to get the correct result.