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Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or )
Given : kf = 1.86
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907
)
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
Answer:
2,2,4-trimethylpentane has the higher heat of combustion in kcal/mol.
Explanation:
Heat of combustion is defined as heat evolved on combustion of 1 mole of a substance.
2,2,4-trimethylpentane has the higher heat of combustion in kcal/mol.
Heat combustion of 2,2,4-trimethylpentane = (1304 kcal/mol) and
1 mole of 2,2,4-trimethylpentane on combustion gives 8 mole sof carbon dioxide gas and 9 moles of water.
Heat combustion of ethanol = (327 kcal/mol)
1 mole of ethanol on combustion gives 2 moles of carbon dioxide gas and 3 mole of water.
Heat combustion of ethanol < Heat combustion of 2,2,4-trimethylpentane