Question

To completely neutralise 200cm3 of 0.5mol/dm3 sodium hydroxide (NaOH), a student adds 100cm3 of 0.5mol/dm3 sulfuric acid (H2SO4). The temperature of the solution goes up by 4.5°C.
The equation for the reaction is



2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)



a. Calculate the amount, in moles, of NaOH in the sodium hydroxide solution.

Answer 1

Answer:

0.1 mol

Explanation:

Step 1: Given data

• Volume of the solution of sodium hydroxide (V): 200 cm³

• Molar concentration of sodium hydroxide (C): 0.5 mol/dm³

Step 2: Convert "V" to dm³

We will use the conversion factor 1 dm³ = 1000 cm³.

200 cm³ × 1 dm³/1000 cm³ = 0.200 dm³

Step 3: Calculate the moles (n) of NaOH

The molarity of the NaOH solution is equal to the moles of NaOH divided by the volume of solution.

C = n/V

n = C × V

n = 0.5 mol/dm³ × 0.200 dm³ = 0.1 mol