Question

20g of solid sodium hydroxide were dissolved in distilled water and made to 400cm³. 30cm³ of this solution required 27cm³ of dilute sulphuric (iv) acid for complete reaction. [Na=23 O=16 H=1]
Determine

1. Moles of sodium hydroxide contained in 30cm³ of solution 
2. Moles of sulphuric (iv) acid that reacted 
3. Concentration of the sulphuric (iv) acid in moles per litre

Answer 1

1. Molar mass of NaOH= 40g
   ²⁰/₄₀ = 0.5M
   0.5 = 400cm³ 
     ? = 1000cm³
   0.5 × 1000
       400
   =1.25M
   1.25 = 1000cm³ 
     ?    =30cm³ 
    1.25 × 30
         1000
   =0.0375moles
2. 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)
   Mol of H₂SO₄ = Mol NaOH Mole ration in equation above is 1:2
                                      2 
                       =0.0375
                             2
   Mol of H₂SO₄ =0.01875moles
3. 0.01875 = 27cm³
      ?        =1000cm³
   1000 × 0.01875 
           27
   =0.6944M