Asked by Anonymous

What mass of CaCl2(s) must be dissolved in pure water at 10 degrees celsius to make a 26.3 mL solution and to increase the solution temperature to 16.4 degrees celcius?

assume solution has same physical properties as water.

deltaH of solution = -82.8 kJ/mol
specific heat of water = 4.184J/g degrees celsius
density of water = 1.00 g/cm^3

DOES ANYONE HAVE ANY CLUE AS 2 HOW 2 START THIS?

any help will b appreciated
Answered by DrBob222
How much heat must be added to the water to move the T from Iinitial = 10 C to Tfinal = 16.4 C. That will be
q = massH2O x specific heat water x (Tfinal-Tinitial) = ??
Knowing that 1 mol CaCl2 dissolving in water will produce 82.8 kJ, how many mols CaCl2 are required. Then convert that to grams. Post your work if you get stuck.
Answered by Anonymous
delta T = 6.4 degrees celsius

q = 26.4g x (4.184J/g*degrees celsius) x 6.4 deg celsius = 706 J = 0.706 kJ

82.8 kJ/mol/ 0.706kJ = 117 mol

(117mol)(111g/mol) = 1.3 x 10^4 g

this doesn't ssem right...did i do something wrong?

and would i use 2 significant digits in the final answer because of the temperature?
Answered by Anonymous
oh never mind...i was supposed to take 0.706kJ and divide by 82.8kJ and the answer would be 0.95g, which makes a lot more sense.

Answered by DrBob222
Right. My answer was 0.944 grams. As to the number of s.f., I assume the 10 C is really 10.0 C. If so, then there are 3 s.f. figures in delta T and the other values; therefore, 3 s.f. are justified. By the way, the mass of water is 26.3 instead of 26.4 and the molar mass of CaCl2 is closer to 111 than it is to 117. You might want to calculate those parts again and see if you get my value.
Answered by Anonymous
693