Answer:
m = 98.6 g
Explanation:
Given:
L = 500mL = 0.5 L
Molarity (M) = 3.4 M
Find:
Amount of grams of Sodium Chloride (NaCl) present in the solution = ?
Solution:
Weight of the solute = Molarity (M) x Volume (in Liters) x Molecular Weight (Molecular Mass)
Determining the Molecular Weight(Molecular Mass) of NaCl:
How many atoms of Na (Sodium) are present in the chemical formula of NaCl? The answer is 1, since it has no subscript.
Na has a Molar mass of 23 g/mol
How many atoms of Cl (Chlorine) are present in the chemical formula of NaCl? The answer is 1, since it has no subscript.
CL has a Molar mass of 35 g/mol
[tex]MM = 23 (1) + 35(1)\\MM = 58 \frac{g}{mol}[/tex]
Applying now on the given formula:
Weight of the solute = Molarity (M) x Volume (in Liters) x Molecular Mass
Weight of the solute = [tex]3.4 * 0.5 * 58\\[/tex]
Weight of the solute = [tex]98.6 g[/tex]
