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Explanation:
<u>Answer:</u> The boiling point of solution is 100.62°C
<u>Explanation:</u>
Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.
The equation used to calculate elevation in boiling point follows:
![\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}](https://tex.z-dn.net/?f=%5CDelta%20T_b%3D%5Ctext%7BBoiling%20point%20of%20solution%7D-%5Ctext%7BBoiling%20point%20of%20pure%20solution%7D)
To calculate the elevation in boiling point, we use the equation:
![\Delta T_b=iK_bm](https://tex.z-dn.net/?f=%5CDelta%20T_b%3DiK_bm)
Or,
![\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}](https://tex.z-dn.net/?f=%5Ctext%7BBoiling%20point%20of%20solution%7D-%5Ctext%7BBoiling%20point%20of%20pure%20solution%7D%3Di%5Ctimes%20K_b%5Ctimes%20%5Cfrac%7Bm_%7Bsolute%7D%5Ctimes%201000%7D%7BM_%7Bsolute%7D%5Ctimes%20W_%7Bsolvent%7D%5Ctext%7B%20in%20grams%7D%7D)
where,
Boiling point of pure water = 100°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal boiling point elevation constant = 0.512°C/m.g
= Given mass of solute = 23.6 g
= Molar mass of solute = 103.5 g/mol
= Mass of solvent (water) = 188 g
Putting values in above equation, we get:
![\text{Boiling point of solution}-100=1\times 0.512^oC/m\times \frac{23.6\times 1000}{103.5g/mol\times 188}\\\\\text{Boiling point of solution}=100.62^oC](https://tex.z-dn.net/?f=%5Ctext%7BBoiling%20point%20of%20solution%7D-100%3D1%5Ctimes%200.512%5EoC%2Fm%5Ctimes%20%5Cfrac%7B23.6%5Ctimes%201000%7D%7B103.5g%2Fmol%5Ctimes%20188%7D%5C%5C%5C%5C%5Ctext%7BBoiling%20point%20of%20solution%7D%3D100.62%5EoC)
Hence, the boiling point of solution is 100.62°C
Answer:
Heat of solution= - 63KJ/mole
Heat evolved = - 18.82KJ
Explanation:Heat of solution = Heat of solute + Heat of hydration
◇Hsoln = - ( - 7.3 × 10 ^ 2) + ( - 793 ) KJ / mol
◇H soln= - 63 KJ
Use enthalpy of solution equation to determine the enthalpy of solution for LiI
LiI ---> Li^ + + I^-
Molecular mass of LiI=133.85g
The chemical equation for the dissociation will be:
40gLiI (1molLiI)/133.85g LiI - 63KJ/1 mol LiI
18.82KJ
The heat was evolved because the value of enthalpy was negative
In setting up your equipment in front of the fume hood to eliminate
any vapors during the experiment. Use an iron ring stand with iron ring to back
up the clay triangle to clasp the crucible and lid. Alter the height of the
iron ring so the end of the crucible is about one centimeter above the inner
blue cone of the strong flame.
1.58 × 10^(-10) in
<em>Step 1</em>. Convert picometres to metres
4.01 pm × [10^(-12) pm]/[1 m] = 4.01 × 10^(-12) m
<em>Step 2</em>. Convert metres to inches
4.01 × 10^(-12) m × 39.37 in/1 m = 1.58 × 10^(-10) in