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BlackZzzverrR [31]
3 years ago
12

a solution is prepared by dissolving 55g of CaCl2 into 300g of water. if the Kf of water is -1.86 c/m, shat is the freezing poin

t depression of the solution
Chemistry
1 answer:
Zepler [3.9K]3 years ago
3 0

Answer:

Relation between , molality and temperature is as follows.

                T =

It is also known as depression between freezing point where, i is the Van't Hoff factor.

Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.

         i for  = 3

          i for glucose = 1

          i for NaCl = 2

Depression in freezing point will have a negative sign. Therefore, d

depression in freezing point for the given species is as follows.

       

                 =

       

                  =

     

                   =

Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.

                    Glucose < NaCl <

Explanation:

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A perfectly spherical balloon occupied 35°C of neon gas under a pressure of 2 atm.
lutik1710 [3]

Answer:

V₂ =  116126.75 cm³

Explanation:

Given data:

Radius of balloon = 15 cm

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Initial temperature = 35 °C (35 +273 = 308K)

Final temperature = -20°C (-20+273 = 253 K)

Final pressure = 0.3 atm

Final volume = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

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V = 4/3πr³

V = 4/3×22/7×(15cm)³

V = 14137.17 cm³

V₂ = P₁V₁ T₂/ T₁ P₂  

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V₂ = 7153408.02 atm .cm³. K / 61.6 K.atm

V₂ =  116126.75 cm³

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