Question

How many grams of NaCl are required to prepare 985 mL of 0.77M NaCl solution?

Answer 1

Attempt this question with the concept of molarity as the question provides us the information about the solution's molar concentration.

Formula used: We will use molar concentration (molarity) formula which is as follows:-
$M=\dfrac{n}{V}$

Complete step-by-step answer: We know that molarity is defined as a number of moles of solute per liter of solution. In other words, it should be said as a measure of the concentration of a chemical species, generally a solute in a solution, in terms of the amount of the substance per unit volume of solution.
Mathematically, $M=\dfrac{n}{V}$
Here, M = molarity of the solution or molar concentration
n= moles of solute
V= volume of solution in Liters
We are provided with 0.77M solution, which means 0.77 moles of NaCl in 1L solution. But desired solution is 985 mL which is equivalent to 0.985L (as $985mL\cdot \dfrac{1\text{ }L}{{{10}^{3}}mL}=0.985\text{ }L$ )
which means number of moles of NaCl in 0.985 mL:-
=$0.985L\cdot \dfrac{0.77\text{ }moles\text{ }NaCl}{1L\text{ }}=0.75845\text{ }moles\text{ }NaCl$
The molar mass of sodium chloride is $58.44gmo{{l}^{-1}}$ which also implies that 1 mole of sodium chloride has a mass of 58.44g. Since we have $0.75845\text{ }moles\text{ }NaCl$, so the mass of NaCl is equivalent to:-
$0.75845\text{ }moles\cdot \dfrac{58.44\text{ }g}{1\text{ }mole\text{ }}=44g$
Therefore, 44 grams of NaCl is required to prepare 985 mL of 0.77M NaCl solution

Note: We can also use the following formula to calculate the mass of NaCl:-
$M=\dfrac{\text{mass of NaCl in gram}}{\text{Molar mass of NaCl}}\times \dfrac{1000}{V(in\text{ mL})}$
$0.77M=\dfrac{\text{mass of NaCl in gram}}{58.44g}\times \dfrac{1000}{985mL}$
Mass of NaCl = 44 grams.
-Kindly remember to accordingly convert units and use them in the formula as required to get the correct result.