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Harman [31]
2 years ago
7

List the following aqueous solutions in order of decreasing boiling point. 0.120 mm glucose, 0.050 m LiBrm LiBr, 0.050 m Zn(NO3)

2m Zn(NO3)2 . Rank items in order of decreasing boiling point.
Chemistry
1 answer:
leva [86]2 years ago
7 0

Answer:

0.050 m LiBrm LiBr < 0.120 mm glucose <0.050 m Zn(NO3)2m Zn(NO3)2

Explanation:

The above aqueous solutions show that LiBr low boiling point followed by glucose and Zinc.

using the equation of boiling point elevation

ΔTb = i×Kb×M

Making temperature constant at a value of 290K you can simple do a calculation to conform the boiling point.

looking the van hoff values for Znc = 3 , LiBr = 2 and glucose = 1

(glucose)ΔTb = i×Kb×M = 1×290k×0.120 = 34.8

(Zinc) = 3×290×0.050 = 43.5

(LiBr)  = 2×290×0.050 =29

∴ in conclusion LiBr has a the lowest decreasing boiling point

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Explanation:

The given data is as follows.

         V_{1} = 85.0 ml,        M_{1} = 0.9 M

         V_{2} = 85.0 ml,        M_{1} = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So,     No. of moles = Molarity × Volume

                                = 0.9 M \times 0.085 L        (as 1 L = 1000 ml so, 85 ml = 0.085 L)

                                = 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

              56.2 kJ/mol \times 0.076 mol

                    = 4.271 kJ

or,                 = 4271 J     (as 1 kJ = 1000 J)

Therefore,    heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence,     mass of HCl = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Similarly,    mass of KOH = = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Hence, total mass of the solution = 85 g + 85 g

                                                        = 170 g

Also,                   q = mC \Delta T

                     4271 J = 170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C    

                     0.0773 = T_{f} - 18.24

                    T_{f} = 18.317^{o}C  

Thus, we can conclude that final temperature of the mixed solution is 18.317^{o}C.

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