Question

Caluclate M, N , Molality w % , v % mole fraction density of solute ... when 5g of Nacl is dissolve in 1000 ml of ..... water density of solution = 0.997 g mol-1

Answer 1

Moles of solute = 5 / 58.44 
                         = 0.085 mol 

volume of solution= 1008 ml 
                             = 1008 / 1000
Molarity = 0.085 / 1.008 
              = 0.84 M 

n factor = N/ M

as we know the n factor for a salt is equal to the total no. positive charge on dissociation   here Na+    ..... value of n factor =  1 

Hence 1 = N /M 

N = M 

 Normality = 0.84  N 

mass of sovent = 1000 g 
                           = 1 kg 

Molality = 0.085 /  1 
 molality = 0.085 

w /w % =  5 / 1005 x 100 

              = 500 / 1005  

              = 0 .49 %



v/ v % = volume of solute / volume of solution 

           = 1000 / 1008 x 100 
              = 99.2 % 
density  of solute = mass / volume 

                  = 5 / 8 = 0.625 g /ml 
                         

moles of water = 1000 / 18 
                      = 55.55

 now mole fraction =  0.085 / 0.085 + 55.55
                              = 0.0015 
 may it helps you ^ _ ^ 

Answer 2

Explanation:

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The above answer is correct mark it as brainliest