Question

Calculate the amount of benzoic acid ${C_6}H{ _5}COOH$ required for preparation of $250mL$ of $0.15M$solution in methanol.

Answer 1

Hint – Start the solution by clarifying the concept behind molarity and moles to make understanding the calculations easier. Then use the formula for molarity,$M = \dfrac{{Number{\text{ }}of{\text{ }}moles}}{{Total{\text{ }}volume{\text{ }}of{\text{ }}the{\text{ }}solution}}$ to find the number of moles $1mole$ in $1liter$ of the solution. Then use this value to determine the number of moles in $250mL$ of the solution. Then define the weight of benzoic acid (weighs about$122.12\;g/mol$) and use this to calculate the amount of benzoic acid in $\dfrac{1}{4}L$ of the given solution.

Complete step by step solution:

Before going on to the mathematical calculations, let’s look at the concept of moles and molarity first.
So, 1 mole is the amount of a substance that has exactly $6.02214076 \times {10^{23}}$atoms or particles of that substance.
- Molarity ($M$) – It is a measurement of concentration of a certain substance in a solution.
- Molarity is by definition the moles of a solute that is present in a liter of the solution.
- Molarity is called the molar concentration of a solution. The unit of molarity is $moles/L$
Let $n$ be the no. of moles in the solution.
Now we know the equation of molarity for $1liter$ of the benzoic acid ${C_6}H{ _5}COOH$ solution is
$M = \dfrac{{Number{\text{ }}of{\text{ }}moles}}{{Total{\text{ }}volume{\text{ }}of{\text{ }}the{\text{ }}solution}}$
$\Rightarrow 0.15 = \dfrac{{Number{\text{ }}of{\text{ }}moles}}{1}$
Now for $250mL$ i.e. $\dfrac{1}{4}L$, the equation becomes
$\Rightarrow 0.15 = \dfrac{n}{1} \times \dfrac{4}{1}$
$\Rightarrow n = \dfrac{{0.15}}{4}moles$
So, we know that $\dfrac{{0.15}}{4}moles$ of benzoic acid (${C_6}H{ _5}COOH$) is required to make $250mL$ of $0.15M$ solution in methanol.
We know by the concept of moles, that $1mole$ of benzoic acid (${C_6}H{ _5}COOH$) weighs about $122.12\;g/mol$.
- So the weight of benzoic acid (${C_6}H{ _5}COOH$) required to make $250mL$ of $0.15M$ solution in methanol is
- Weight of benzoic acid (${C_6}H{ _5}COOH$) $= \dfrac{{0.15}}{4} \times 122.12$
- Weight of benzoic acid (${C_6}H{ _5}COOH$) $= 4.57g$

Note – In such types of problems, we consider that all of the solute dissolves completely in the solution but in reality the situation may be different due to various factors like temperature, saturation, pressure, etc. Benzoic acid dissolves readily in methanol (solubility of benzoic acid at $- 18^\circ C$ in methanol is $30g/100g$).