Question

You have 25 mL of acetic acid $(C{H_3}COOH)$ density =1.058 gm/L. How many grams do you have? How many mol do you have? How many molecules do you have? Help please

Answer 1

Hint: This whole question is solved in three steps, first the mass is calculated by using the formula of density where density is equal to mass divided by volume. In the second step the mole is calculated by dividing the mass with the molecular weight. In the third step molecules are calculated by multiplying the moles with the Avagadro’s number.

Complete step by step answer:
Given,
The volume is 25 mL
Density is 1.058 gm/L.
The formula which relates the density and volume is given as density is equal to the mass divided by the volume.
The formula is shown below.
$D = \dfrac{m}{V}$
D is the density
m is the mass
V is the volume.
To calculate the mass, substitute the values in the above equation.
$\Rightarrow 1.058gm/L = \dfrac{m}{{0.025L}}$
$\Rightarrow m = 1.058gm/L \times 0.025L$
$\Rightarrow m = 0.0264g$
Therefore, the mass of acetic acid is 0.0264 g.
The molecular weight of acetic acid $C{H_3}COOH$ is 60.05 g/mol.
The formula to calculate the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles
m is the mass
M is the molecular weight
To calculate the moles of acetic acid, substitute the values in the above equation.
$\Rightarrow n = \dfrac{{0.0264g}}{{60.05g/mol}}$
$\Rightarrow n = 0.00043$mol
Therefore, the number of moles present in acetic acid is 0.00043 mol.
We know that 1 mole of any substance is equal to $6.022 \times {10^{23}}$ atoms where the value is known as the Avagadro’s number and the constant is said as Avagadro’s constant.
So, the number of atoms is calculated as shown below.
$\Rightarrow 0.00043 \times 6.022 \times {10^{23}}$
$\Rightarrow 2.589 \times {10^{20}}$

Therefore, the number of molecules is $2.589 \times {10^{20}}$.

Note: Make sure to convert the value of volume given in milliliters in liters as the density is calculated in terms of grams per liter. 1mL is equal to $\dfrac{1}{{1000}}$L.