Respuesta :
The question is incomplete,the complete question :
Calculate the molality of a 10.0% (by mass) aqueous solution of hydrochloric acid:
a) 0.274 m
b) 2.74 m
c) 3.05 m
d) 4.33 m
e) the density of the solution is needed to solve the problem
Answer:
The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.
Explanation:
10.0% (by mass) aqueous solution of hydrochloric acid.
10 grams of HCl is present in 100 g of solution.
Mass of HCl = 10 g
Mass of solution = 100 g
Mass of solution = Mass of solute + Mass of water
Mass of water = 100 g - 10 g = 90 g
Moles of HCl = [tex]\frac{10 g}{36.5 g/mol}=0.2740 mol[/tex]
Mass of water in kilograms = 0.090 kg
Molality = [tex]\frac{0.2740 mol}{0.090 kg}=3.05 mol/kg[/tex]
The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.
Molality is the measure of concentration of solute in 1 kg of solution. The molality of the solution is 3.05 mol/kg.
10% of HCl (by mass) means 10 g of HCl and in 90 g of water.
Molar mass of HCl = 36.5 g/mol
Molality:
It is the measure of concentration of solute in 1 kg of solution. It can be calculated by the formula.
[tex]\bold {m = \dfrac {n }{w}\times 1000}[/tex]
Where,
m- molality
n - number of moles
w - weight of solvent in grams
Number of moles of HCl
[tex]\bold {n = \dfrac w{m} = \dfrac {10}{36.5} = 0.274 g}[/tex]
put the value in molality formula,
[tex]\bold {m = \dfrac {0.274 }{90}\times 1000}\\\\\bold {m = 3.05\ g/mol}[/tex]
Therefore, the molality of the solution is 3.05 mol/kg.
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