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:
Answer of question a is 345J.
Explanation:
In question a following is given in data:
-mass of iron (m) = 10.0 g
-temperature (ΔT) = final temperature- initial temperature= 100-25= 75 degree Celsius
-Specific Heat capacity of iron (c)= 0.46J/g°C.
Heat (Q)=?
Solution:
Formula for Heat is :
Q=m x c x ΔT
Q= 10 x 0.46 x 75
Q= 345 J.
so, 345 joules of heat is needed to increase the temperature of 10 grams of iron.
- From the above formula all other questions can easily be solved from the same procedure.
Xe !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!