2.20g of a mixture of sodium chloride and sodium carbonate were made up to 250cm³ of aqueous solution. 25cm³ of the solution required 24.8cm³ of 0.100 mol/dm³ of hydrochloric acid for neutralisation

Question

Calculate the number of moles of hydrochloric acid which neutralised 25cm³ of the solution?

Answer 1

To find the number of moles of hydrochloric acid (HCl) that neutralized 25 cm³ of the solution, we can use the volume and concentration of the HCl solution.

The formula for calculating moles is:

\[ \text{Moles} = \text{Concentration (mol/dm}^3) \times \text{Volume (dm}^3) \]

1. Convert the volume of HCl from cm³ to dm³: \[ 24.8 , \text{cm}^3 = \frac{24.8}{1000} , \text{dm}^3 = 0.0248 , \text{dm}^3 \]

2. Given concentration of HCl: \[ \text{Concentration} = 0.100 , \text{mol/dm}^3 \]

3. Calculate the number of moles of HCl: \[ \text{Moles of HCl} = 0.100 , \text{mol/dm}^3 \times 0.0248 , \text{dm}^3 = 0.00248 , \text{mol} \]

So, the number of moles of hydrochloric acid that neutralized 25 cm³ of the solution is 0.00248 moles.