Question

You want to prepare a standard solution of 1 mol-dm³ magnesium chloride. The volume of the volumetric flask is 100 ml. How many grams of crystals do you need?​

Answer 1

The answer is: 9.521 grams of crystals are needed to prepare the solution with molarity  $1\ mol/dm^3$.

Molarity of a substance is defined as moles of solute in 1 L of solution .

$Molarity =\frac{Moles\ of\ solute}{Volume\ of\ solution\ in\ L}$

How are moles and mass related?

• Amount of atoms are usually expressed in term of moles. Moles are calculated as- $Moles =\frac{Mass}{Molar\ mass}$
• Thus, mass is determined by multiplying the moles of substance with its molar mass.
• Now, in the given solution of magnesium chloride, molarity = $1\ mol/dm^3$.
• $1\ dm^3 = 1\ L = 1000\ ml$, thus, volume = 0.100 L
• Thus, the moles of magnesium chloride can be determined as-

$1\ mol/L = \frac{Moles\ of\ MgCl_2}{0.100 L}$

$Moles\ of\ MgCl_2 = (1\ mol/L)* (0.1\ L) == 0.1\ moles$

• Next, using the moles and molar mass of $MgCl_2$ (= 95.21 g/mol), the mass of crystals are calculated-

$Mass\ of\ MgCl_2 = (0.1\ moles) * (95.21\ g/mol) = 9.521\ g$

• Therefore, 9.521 grams of crystals are needed to prepare the solution with molarity  $1\ mol/dm^3$.

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