(a)
Interpretation:
Three beakers of same solution ‘A’, ‘B’ and ‘C’ made of water and non-volatile solute are given –
Figure 1
The solution having higher vapor pressure has to be identified.
Concept Introduction:
Vapor pressure of a substance is known as the pressure exerted by molecules on the vapor phase when they are in equilibrium with their actual phase which can be liquid or solid.
A substance is said to be volatile if it vaporizes readily at room temperature itself. Such substances have high vapor pressure as most of its molecules tend to exist in vapor phase. A substance is said to be non-volatile if it doesn’t vaporize spontaneously and remains stable.
Vapor pressure of a volatile solvent can be lowered by addition of a non-volatile solute. Raoult’s law deals with the vapor pressure of pure solvents and solution which states –
Partial pressure of solvent is equivalent to the product of vapor pressure of the solvent in its pure state and mole fraction of solvent in the solution. It is expressed as,
Where,
When the solute is non-volatile, the vapor pressure of the whole solution is equal to
The lowering of vapor pressure of the solvent due to the addition of non-volatile solute is expressed as,
Where,
(b)
Interpretation:
Three beakers of same solution ‘A’, ‘B’ and ‘C’ made of water and non-volatile solute are given –
Figure 1
The solution with lowest boiling point has to be identified.
Concept Introduction:
Boiling point of a liquid substance is defined as the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure.
Boiling point of a substance can be determined by the formula,
Where,
(c)
Interpretation:
Three beakers of same solution ‘A’, ‘B’ and ‘C’ made of water and non-volatile solute are given –
Figure 1
A laboratory procedure to make all these three solutions to have same freezing point has to be described.
Concept Introduction:
Freezing point of the substance is temperature at which liquid substance remains in equilibrium with solid substance.
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Chapter 12 Solutions
Student Solutions Manual for Ebbing/Gammon's General Chemistry, 11th
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- Nonearrow_forwardJON Determine the bund energy for UCI (in kJ/mol Hcl) using me balanced chemical equation and bund energies listed? का (My (9) +36/2(g)-(((3(g) + 3(g) A Hryn = -330. KJ bond energy и-н 432 bond bond C-1413 C=C 839 N-H 391 C=O 1010 S-H 363 б-н 467 02 498 N-N 160 N=N 243 418 C-C 341 C-0 358 C=C C-C 339 N-Br 243 Br-Br C-Br 274 193 614 (-1 214||(=olin (02) 799 C=N 615 AALarrow_forwardDetermine the bond energy for HCI ( in kJ/mol HCI) using he balanced cremiculequecticnand bund energles listed? also c double bond to N is 615, read numbets carefully please!!!! Determine the bund energy for UCI (in kJ/mol cl) using me balanced chemical equation and bund energies listed? 51 (My (9) +312(g)-73(g) + 3(g) =-330. KJ спод bond energy Hryn H-H bond band 432 C-1 413 C=C 839 NH 391 C=O 1010 S-1 343 6-H 02 498 N-N 160 467 N=N C-C 341 CL- 243 418 339 N-Br 243 C-O 358 Br-Br C=C C-Br 274 193 614 (-1 216 (=olin (02) 799 C=N 618arrow_forward
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