Answer:
2,347.8 grams
Explanation:
The freezing point depression Kf of water = 1.86° C / molal.
To still freeze at -12° C, then the molality of the solution 12/ 1.86 = 6.45 moles
The molecular weight of sorbitol (C6H14O6)is:
6 C = 6 ×12 = 72
14 H = 14 × 1 = 14
6 O = 6 × 16 = 96
...giving a total of 182
So one mole of sorbitol has a mass of 182 grams.
Since there are 2 kg of water, 2 × 6.45 moles = 12.9 moles can be added to the water to get the 12° C freezing point depression.
Therefore
grams = moles × molar mass
12.9 moles × 182 grams / mole = 2,347.8 grams of sorbitol can be added and still freeze
This is actually incorrect, the distance as stated above either from trough to trough or from crest to crest within a corresponding wave, would be referred or called as the wavelength of the given wave.
1. Take 100ml of water as solvent and boil it few minutes.
2. Now add one tea spoon sugar, one tea spoon tea leaves and 50ml of milk. Here sugar, tea leaves and milk are solute.
3. Now boil it again for few minutes so that sugar will dissolves in solution as sugar is soluble in water
4. Now filter the solution. Collect the filtrate in cup. The insoluble tea leaves will be left behind as residue.
