Question

2. What is the molarity of a solution made from 325.4g of Aluminum chloride with enough water to
make 500.0 mL? *
(1 Point)
AICI: Al = 26.982 CI = 35.45
mol

Answer 1

Given :

• Mass of aluminium chloride = 325.4g
• Volume of solution = 500ml
• Atomic mass of Al = 26.982
• Atomic mass of Cl = 35.45

To find :

Molarity of given solution

Formula used :

• Number of mole = mass ÷ molar mass
• Molarity = number of mole ÷ volume of solution in L

Solution :

We have to find molar mass of Aluminium chloride ( AlCl₃ )

➝ Molar mass of AlCl₃ = ( 1 × atomic mass of Al) + (3 × atomic mass of Cl )

➝ Molar mass of AlCl₃ = ( 1 × 26.982) + (3 × 35.45 )

➝ Molar mass of AlCl₃ = 26.982 + 106.35

➝ Molar mass of AlCl₃ = 133.332 g/mol

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Now we will find number of mole of AlCl₃

➝ Number of mole of AlCl₃ = mass ÷ molar mass

➝ Number of mole of AlCl₃ = 325.4 ÷ 133.332

➝ Number of mole of AlCl₃ = 2.64 mol

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Now molarity ,

$\sf \implies \: Molarity \: =\: \dfrac{ Number \: of \: mole \: of \: AlCl_3 }{Volume \: of \: solution \: in \: L} \\ \\ \sf \implies \: Molarity \: =\: \dfrac{ 2.64 }{ \frac{500}{1000} } \\ \\ \sf \implies \: Molarity \: =\: \dfrac{2.64 \times 1000}{500} \\ \\ \sf \implies \: Molarity \: =\:2.64 \times 2 \\ \\ \sf \implies \: Molarity \: =\: \bold{5.28 \: M}$

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ANSWER : 5.28 M