Question

0.456 grams of a monoprotic acid is titrated with 45.88 mL of 0.0500 M NaOH. What is the molecular mass (molar mass) of the acid

Answer 1

Answer:

$Molar\ mass= 198.78\ g/mol$

Explanation:

Considering:

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

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For NaOH :

Molarity = 0.0500 M

Volume = 45.88 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 45.88×10⁻³ L

$Moles_{NaOH} =0.0500 \times {45.88\times 10^{-3}}\ moles=0.002294\ moles$

Moles of NaOH = Moles of monoprotic acid

So, moles of monoprotic acid = 0.002294 moles

Given that:- mass = 0.456 g

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.002294\ moles= \frac{0.456\ g}{Molar\ mass}$

$Molar\ mass= 198.78\ g/mol$