As you increase elevation the atmospheric pressure decreases, which means the boiling point of water also decreases. In Denver, the Mile High City, the boiling point of water is 95.0 degree C. a. How many grams of your assigned salt (using the van't Hoff factor you calculated in lab) would need to be added to 1.0 gallon (3.8 kg) of water to raise the temperature to 100.0 degree C, the boiling point of water at sea level. b. In the USA, a volume (cup) is used more commonly in cooking than a weight (grams). Convert the grams of salt you calculated into cups (1 cup = 273 g). c. Ignoring the potential toxicity of your assigned salt, is the amount of salt needed to increase the boiling point of the water 5 degree C reasonable?

Check the explanation

Explanation:

colligative property

a. let the salt in question is NaCl

to raise temperature of water from 95 to 100 oC

dTb = 100-95 = 5 oC

use,

dTb = iKbm = 5 oC

      = 2 x 0.512 x grams NaCl/58.44 x 3.8

grams of NaCl salt needed = 1084.33 g

b. with,

1 cup = 273 g

thus,

amount of salt NaCl = 1084.33 g/273 = 3.97 cups (or 4 cups)

c. Yes, to raise the temperature of water by 5 oC the amount of salt needed for 3.8 kg water is reasonable.

a) The grams of the salt needed to be added is 1084.3 g

b) 4 cups

c) See the explanation.

Explanation:

a) If the salt is NaCl, and the temperature rise from 95 to 100°C is equal to:

dT = 100 - 95 = 5°C

dT = i*Kb*m = 5 = 2 * 0.512/58.44 * 3.8

m = 1084. 3 g of salt needed

b) If 1 cup = 273 g, thus:

m = 1084.3/273 = 3.9 cups = 4 cups

c) Taking into account the amount of salt needed to raise the temperature by 5 ° C, the amount of salt that was calculated is reasonable

1 is the answer to this problem

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2