Question

$ 0.543g $ of $ CaC{l_2} $ was dissolved in enough water to make a $ 500ml $ solution. What is the concentration of $ C{l^ - } $ in $ \% w/v $ ?

Answer 1

Hint: Weight of calcium chloride was given, from molar mass and weight, to determine the moles of calcium chloride. By taking the given volume, and determined moles molar concentration of calcium chloride will be known. From this molar concentration of chloride ions can be calculated. It is useful to write the percentage of weight by volume.

Complete Step By Step Answer:
Calcium chloride is an ionic salt with the molecular formula of $ CaC{l_2} $ , it can dissociate into calcium ions, and two chloride ions.
The number of moles is the ratio of the weight of the compound to the molar mass of that compound.
Molar concentration is also known as molarity which is the ratio of the number of moles to volume of solution in litres.
Given that $ 0.543g $ of $ CaC{l_2} $ was dissolved in enough water to make a $ 500ml $ solution. Molar mass of $ CaC{l_2} $ is $ 110.98gmo{l^{ - 1}} $
Number of moles of $ CaC{l_2} $ will be $ \dfrac{{0.543}}{{110.98}} = 0.004893moles $
Substitute these moles to get molarity which will be $ \dfrac{{0.00489}}{{500}} \times 1000 = 0.009786M $
As calcium chloride dissociates into two chloride ions.
Molar concentration of $ C{l^ - } $ will be $ 2 \times 0.009786 = 0.01957M $
As $ \% w/v $ has to be determined, substitute the molarity, volume as $ 100ml $ and molar mass as $ 35.5gmo{l^{ - 1}} $
 $ \% w/v = \dfrac{{0.01957 \times 100 \times 35.5}}{{1000}} = 0.06938g $
Thus, $ 0.543g $ of $ CaC{l_2} $ was dissolved in enough water to make a $ 500ml $ solution then the concentration of $ C{l^ - } $ in $ \% w/v $ will be $ 0.06938g\% $ .

Note:
While calculating the molarity, the volume of solution must be taken in litres only. As the definition indicates the moles of solute in volume of solution in litres. If it is taken in millilitres, the molarity should be multiplied with $ 1000ml $ as one litre is equal to $ 1000ml $ .