What is the molality of a solution containing 25.03 g FeCl3 in 5.00 L of pure water? (molar mass of FeCl3=
162.20 g/mol).
a. 0.30 m b. 0.03 m
c. 0.03086 m
d. 0.0309 m


Sagot :

The molality of a solution : 0.03 m(b)

Further explanation  

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. Concentration shows the amount of solute in a unit of the amount of solvent.  

Molality shows how many moles are dissolved in every 1000 grams of solvent.  

[tex]\tt molality(m)=\dfrac{mol~solute}{kg~solvent}=\dfrac{mol~solute\times 1000}{mass~solvent(g)}[/tex]

m = Molality  

n = number of moles of solute  

p = solvent mass (grams)  

Given

25.03 g FeCl3 in 5.00 L of pure water

Required

The molality

Solution

mol solute = mol FeCl3 :

= mass : molar mass

= 25.03 g : 162.20 g/mol

= 0.1543

mass solvent=mass water :

= 5000 ml(5 L) x 1 g/ml

= 5000 g = 5 kg

The molality :

= 0.1543 mol : 5 kg( 4 sig fig : 1 sig fig)

= 0.03 m(the result must be 1 sig fig)

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