6. Given the solubility chart for salts from your class notes (ppt), determine w

Question

6. Given the solubility chart for salts from your class notes (ppt), determine whether each solution listed is saturated, unsaturated, or supersaturated.

30g KCl in 100g water at 40C

25g Pb(NO3)2 in 50g water at 5C

160g NaNO3 in 200g water at 10C

7. What is the molality of a solution containing 85.0g NaCl in 2.0 kg water (density of water = 1g/mL)? How would this value compare to the molarity of this solution? What is the utility of molality as a concentration unit (why do we use it)? When do we use molality rather than molarity?

Explanation / Answer

KCl solubility in water at 40 degrees is 42.6 g/100 g

so its an unsaturated solution

lead nitrate in 100 g water

at 0 degree= 37.5 g

aat 10 degree=46.2 g

so at 5 degree, it must be between these two values.

in 50 g, the value will be half, around 20 g.

So, 25 g will make supersaturated solution.

NaNO3:

at 10 degree, in 100 g, solubility= 60.8 g

in 200 g, its approximately 160 g.

So it makes a saturated solution

7.

molality= no. of moles of solute/kilogram of solvent

molar mass of NaCl= 58.44g

moles of NaCl= 85/58.44 = 1.45

molality= 1.45/2= 0.7273 moles/Kg

to calculate molarity, first covert Kg to litres:

2 kg= 2 litres.

so molality= molarity= 0.7273

Molalities are more convenient than molarities in experiments that involve significant temperature changes. Because the volume of a solution increases when its temperature increases, heating makes the solutions molarity go down- but the molality, which is based on masses rather than volumes, remains unchanged.

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