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andreev551 [17]
3 years ago
5

Logan is cooking, and he wants to increase the boiling point of the water. Which solute would not be a good choice as it would n

ot increase the boiling point as much as the others?
A. Salt, NaCl

B. Magnesium Sulfate, MgSO4

C. Potassium Sulfate, K2SO4

D. Sucrose, C12H22O11
Chemistry
1 answer:
Mars2501 [29]3 years ago
8 0
Sugar would not increase the boiling point of the water as much as the others.To show this, we recall the equation for boiling point elevation ∆t:  
     ∆t = i Kb m 

The difference in these four solutions will be in the product of the van't Hoff Factor i and molality m: i * m. 

If we assume that we have 0.1m of each solute, the products i*m for these solutions are:
     
     Solution                      Identity of particles                          i      i * m
        0.1 m NaCl                 Na+, Cl-                                         2     2 * 0.1 = 0.2
        0.1 m MgSO4             Mg^2+, SO4^2-                            2     2 * 0.1 = 0.2
        0.1 m K2SO4              two K+, SO4^2-                           3     3 * 0.1 = 0.3
        0.1 m C12H22O11      covalently bonded molecules     1     1 * 0.1 = 0.1

Therefore, having the lowest  i * m, sucrose will have the lowest boiling point elevation.
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Answer: The molar enthalpy change is 73.04 kJ/mol

Explanation:

HCl+NaOH\rightarrow NaCl+H_2O

moles of HCl= molarity\times {\text {vol in L}}=0.415mol/L\times 0.1=0.0415mol

As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.

volume of water = 100.0 ml + 50.0 ml = 150.0 ml

density of water = 1.0 g/ml

mass of water = volume \times density=150.0ml\times 1.0g/ml=150.0g

q=m\times c\times \Delta T

q = heat released

m = mass  = 150.0 g

c = specific heat = 4.184J/g^0C

\Delta T = change in temperature = 4.83^0C

q=150.0\times 4.184\times 4.83

q=3031.3J

Thus 0.0415 mol of HCl produces heat = 3031.3 J

1 mol of HCL produces heat = \frac{3031.3}{0.0415}\times 1=73043.3J=73.04kJ

Thus molar enthalpy change is 73.04 kJ/mol

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