The volume occupied by 0.25 mol of sulfur dioxide at rtp
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Given :
A 250 ml beaker weighs 13.473 g .
The same beaker plus 2.2 ml of water weighs 15.346 g.
To Find :
How much does the 2.2 ml of water, alone, weigh .
Solution :
Now, mass of water is given by :
Therefore , mass of 2.2 ml of water alone is 1.873 g .
Hence , this is the required solution .
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