First of all, I too have diped my hand in liquid nitrogen. As have all of my highschool friends (chemistry and physics classes made us do it). As long as you do it quickly, you never feel a thing - the nitrogen boils from your hand heat and you never actually touch any liquid nitrogen, even if you put your hand all the way down to the bottom of the container. If you do it slowly, you will lose you hand.
Back to the main subject:
A lot of misleading information in this thread (with a PhD in chemical engineering, I better know this stuff). Even one of those
links mentioned this fact: "Please note that [everything in this link] will not hold true if you take two identical pots containing one gallon of water each and add the salt to one pot".
Back to the original topic, there are two possible cases:
1) You take 1 L of water, and compare it to 1 L of water + 433 gm of salt.
2) You take 1 L of water, and compare it to 0.8 L of water + 433 gm of salt.
Leolaw clearly asked about case #1 in the original post, but you all answered about case #2 - meaning you didn't answer the question that was asked. Leolaw, do you clearly understand the difference of these two experiments?
Lets keep things simple and assume the properties of materials are constant with temperature. Lets also assume everything starts at 20°C. Thus the heat capacity of water is 4.182 J/g°C and the heat capacity of salt (assumed to be NaCl) is 0.86 J/g°C. The density of salt is 2.165 g/cm^3, thus 200 mL of salt weighs 433 g.
If you have 1 L of water (1000 g), you need 4.182 J/g°C * 1000 g * (100°C - 20°C) = 3.346*10^5 J of energy to reach 100°C.
If you have 433 g of salt, you need 0.86 J/g°C * 433 g * (100°C - 20°C) = 2.979*10^4 J of energy to heat it up to 100°C. Thus if you combine both of those (1 L of water + 433 g of salt), you need 3.644*10^5 J of energy to reach 100°C. Note how this is more energy required than with just pure water. Meaning it will automatically take 8.9% longer to reach 100°C (assuming you add heat at the same rate)! But in this salt water mix, you still haven't reached its boiling point - meaning you still need to add even more heat to get it to boil (and thus more time).
A colligative properly means that it doesn't matter what you add, but only the amount that you add. Thus 1 M of salt water has the same boiling point a 1 M of sugar water. This rule isn't always held 100% perfectly, but for an first assumption it is quite close to reality. The boiling point of water is: 100°C + Cm * 0.521°C/M. Where Cm is the molar concentration of whatever was added to water.
The molecular weight of NaCl is 54.88 g/mol. Thus you added 433 g / 58.44 g/mol = 7.409 mol of salt. Hopefully you can calculate the molar concentration of your 1L water + 7.409 mol salt mixture. Then multiply that result by 0.521°C/M and add 100°C. That will give you the boiling point for your mixture. I'll let you do the rest of your homework to see how much total heat is needed to reach this new boiling point. Hint: it will be more heat than the 3.644*10^5 J I mentioned above.
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