Respuesta :
Answer:
12.82 mol/L the molarity of the HCl.
Explanation:
Suppose in 100 grams of 39.0% (by mass) HCl in water.
Volume of solution = V
Density of the solution = d = 1.20 g/mL
Mass = Density × Volume
[tex]V=\frac{M}{d}=\frac{100 g}{1.20 g/mL}=83.33 mL = 0.08333 L[/tex]
Mass of HCl = 39.0% of 100 grams= [tex]\frac{39}{100}\times 100g=39 g[/tex]
Moles of HCl = [tex]\frac{39 g}{36.5 g/mol}=1.0685 mol[/tex]
[tex]Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution (L)}}[/tex]
The molarity of the HCl = M
[tex]M=\frac{1.0685 mol}{0.0833 L}=12.82 mol/L[/tex]
12.82 mol/L the molarity of the HCl.
Answer:
The commercial grade of HCl solution having a density of 1.20 g/ml has the molarity of 12.8 M.
Explanation:
Let's take the volume of solution to be 1000 ml.
Mass of HCl = [tex]\rm density\;\times\;volume[/tex]
Mass = [tex]\rm 1.20\;\times\;1000[/tex]
Mass of HCl = 1200 g.
In 39 % of HCl,
mass of HCl = [tex]\rm 39\;\times\;1200[/tex]
mass of HCl = 468 grams.
Molarity = [tex]\rm \frac{moles}{Liter}[/tex]
Molarity = [tex]\rm \frac{468}{36.5}\;\times\;\frac{1}{liters}[/tex]
Moalrity = 12.8 moles/liter
Molarity of 39% HCl with a density of 1.20 g/ml is 12.8 M.
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