Please give me explanation for every questions. I really want to know how it is

Question

Please give me explanation for every questions. I really want to know how it is been solved but not just the answers.

When HNO2 is dissolved in water it partially dissociates according to the equation HNO2 H+ + NO -2 . A solution is prepared that contains 4.000g of HNO2 in 1.000 kg of water. Its freezing point is found to be -0.1692degree C. Calculate the fraction of HN02 that has dissociated. Express your answer using two significant figures. A 50.0-mL solution is initially 1.58% MgCL, by mass and has a density of 1.05 g/mL. What is the freezing point of the solution after you add an additional 1,35g MgCL,? (Use i = 2.5 for MgCl2 ) Express your answer using two significant figures.

Explanation / Answer

12.114)

Moles HNO2 = 4 g / 47.0147 g/mol = 0.085
m = moles / Kg = 0.085 / 1.000 = 0.085
kf = 1.86
0.1692 = 0.085 x 1.86 x i
i = 1.07
Remember that if i = 2 the dissociation is 100%

12.124)

( 50.0 mL) (1.05 g/mL) = 52.5 grams of solution

(1.58% by mass) (52.5 grams of solution) = 0.8295 grams of MgCl2

52.5 grams of solution - 0.8295 grams of MgCl2 = 51.6705 g H2O

0.8295 grams of MgCl2 + an additional 1.35 g MgCl2 = 2.1295 grams of MgCl2 total

using molar mass, find moles
(2.1295 grams of MgCl2) (1 mole MgCl2 / 95.21 grams) = 0.0223366 moles MgCl2

==================================


52.5 grams of solution - 0.8295 grams of MgCl2 = 51.67 g H2O

the molality is
( 0.022366 moles MgCl2) / 0.05167 kg water = 0.4329 molal


================================

dT = i Kf (molality)

dT = 2.5 (1.86 C / molal) ( 0.4329 molal)

dT = 2.01 C drop in temp

the new freezing point will be
- 2.0 Celsius

12.40)

Potassium nitrate has a lattice energy of -163.8 kcal/mol and a heat of hydration of -155.5 kcal/mol.

1 J = 4.19 cal
1 kJ = 4.19 Kcal
111 kJ = 464.7 kcal

lattice energy of -163.8 kcal/mol is required to break the bonds in the lattice

heat of hydration of -155.5 kcal/mol.is released as the KNO3 is hydrated and becomes more stable

Since lattice energy is required and energy of hydration is released, the net result is 8.3 Kcal of heat energy is required to dissolve a mole of KNO3 in 200 moles of water.


163.8 - 155.5 = 8.3Kcal / mole

8.3Kcal / mole * x moles = 464.7 kcal


Moles = 56 moles of KNO3
Molar mass of KNO3 = 39.1 + 14 +(3 * 16) = 101.1g/mole

101.1g/mole * 56 moles =5661.6 g of KNO3

5661.6 g of KNO3 = 5.6616 Kg of KNO3

12.102)

lattice energy means this.
Na+ + OH- ----> NaOH.. .dHrxn = -887 kJ/mole

heat of hydration means this
Na+ + OH- + mH2O ----> infinitely dilution solution of Na+ + OH-....dHhyd = -932 kJ/mole

********
so to find the heat of dissolution...

NaOH ---> Na+ + OH-...dHrxn = +887 kJ/mole
Na+ + OH- + mH2O ---> dilution... dHrxn = -932 kJ/mole
_______________________________
add...
NaOH ----> diluted Na+ + OH- solution... dHrxn = -45 kJ/mole

and you are correct to here.

*********************
then...from a heat balance
heat evolved by dissolution of NaOH = heat gained by solution

then...
(m x dHdissolution) = (m Cp dT) solution
26.5g x (1 mole / 40.00g) x (45 kJ/mole) = X mL x (1.05g / mL) x (0.0040 kJ/gC) x (100.0

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