concentration of solution

No. of moles of NaOH used = 4/(23 + 16 + 1) = 0.1 mol

In chemistry calculations, we often make simplification to assume that thevolume of final solution is equal to the volume of water added. Actually, this is not true. When 4 g of NaOH is dissolved in 0.5 dm³ of water, the volume of solution formed is slightly greater than0.5 dm³. Therefore,

The volume of solution formed = (0.5 + ΔV) dm³
where ΔV > 0 but is small.

The concentration of solution = 0.1/(0.5 + ΔV) M
Since ΔV is a small positivenumber, the actual concentration is slightly lower than 0.1/0.5 M, i.e. 0.2 M.

2012-03-19 03:30:50 補充：
ΔV is small, so that the concentration is SLIGHTLY lower than 0.2 M.

Solid NaOH is hygroscopic, but 0.2 M NaOH solution is NOT.

NaOH本身係hygroscopic的,佢係solid form既時侯會吸空氣入面既moisture,所以你磅果4g既NaOH入面會含有水份,真正得到既NaOH只係得3.幾g,溶左入水都係會lower than 0.2M

2012-03-18 17:58:28 補充：
the concentration of NaOH would still be 0.2M in the above case,even though it doesn't completely dissociate
10M CH3COOH is still 10M even though not more than 1M of CH3COOH ionize!

2012-03-18 20:34:19 補充：
補充係答樓上果度問既野,答案係低於0.2M

2012-03-18 20:36:57 補充：
umm...我覺得NaOH既hygroscopic property先係造成呢個影響既主因...樓下講得可能冇錯,但個effect實際有幾大?係呢個case ΔV大概會有幾大?

I am not sure about this, but I am thinking whether the ionization of NaOH in water would lower the concentration of the solution.

NaOH + H2O --> Na+ + OH- + H2O