Question

How many grams of MgCl2 remains after 367.2 mL of a 1.24M solution is evaporated to dryness?

Answer 1

Step 1

Molarity is defined as:

`Molarity\text{ \lparen mol/L or M\rparen = }\frac{number\text{ of moles of solute }}{Volume\text{ of solution \lparen L\rparen}}=(n)/(V(L))`

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Step 2

Data provided:

Molarity = 1.24 M

Volume = 367.2 mL

(1 L = 1000 mL => V = 367.2 mL = 0.3672 L)

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Data needed:

The molar mass of MgCl2 = 95.2 g/mol

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Step 3

Procedure:

Molarity = n/V(L)

Molarity x V(L) = n

1.24 M x 0.3672 L = n

0.46 moles = n

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Now, n = mass of MgCl2/molar mass MgCl2

n x molar mass of MgCl2 = mass of MgCl2

0.46 moles x 95.2 g/mol = mass of MgCl2

43.8 g = mass of MgCl2

Answer: 43.8 g of MgCl2 remains after evaporating