Question

20 grams of NAOH is completely neutralized by 19.6 grams of sulphuric acid q then the percent purity of NAOH sample is

Answer 1

Answer:

79.9 %

Explanation:

Moles of Sulfuric acid :

Given, Mass of water = 19.6 g

Molar mass of water = 98.079 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{19.6\ g}{98.079\ g/mol}$

$Moles\ of\ Sulfuric\ acid= 0.1998\ mol$

The reaction between NaOH and H₂SO₄ is shown below as:

2NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2H₂O

1 mole of H₂SO₄ react with 2 moles of NaOH

0.1998 mole of H₂SO₄ react with 2*0.1998 moles of NaOH

Moles of NaOH = 0.3996 moles

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$0.3996\ Moles= \frac{Mass}{39.997\ g/mol}$

Mass of NaOH = 15.98 g

Mass given = 20 grams

% Purity = ( 15.98 )/20 *100 = 79.9 %