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: I think the right answer is c
Answer:
x = 2+
Explanation:
1) FADH2 + Q => FAD + QH2
Since H is added to Q
=> Reactant reduced is Q
(2) Balancing charges on both sides of the equation gives:
QH2 + 2 cyt c(Fe3+) => Q + 2 cyt c(Fe2+) + 2 H+
Thus x = 2+
Answer: 2-Benzyloxyethanol is produced reaction in the fig. below.
Explanation: