Question

250 ml of a sodium carbonate solution
contains 2.65 gram kof Na2CO3. If 10 ml of
this solution is diluted to one litre, what is
the concentration of the resultant solution?
(mol. wt. of Na2CO3 = 106)
(a) 0.1 M (b) 0.001 M
(c) 0.01 M (d) 104
M

Answer 1

heya.....



Concentration of a solution can be expressed in terms of Molarity. Molarity can be calculated in the following way: 


Molarity = moles of solute/ liters of solution

Find moles of 2.65 g of Na₂CO₃ (molar mass = 106)


Moles = mass/molar mass


           = 2.65 g / 106       
    = 0.025 moles


Molarity of this solution: 


Molarity = 0.025 moles / 0.25 liters (250ml in liters)   
           = 0.1 M


We can use the molarity calculated to find the moles of Na₂CO₃ in 10ml solution:


If molarity = moles of solute/ volume in litersThen ; 
  Moles  = volume in liters x Molarity


Moles of Na₂CO₃ in 10ml solution  = 10/1000 × 0.1M                                                           = 0.001 moles


We can find the new molarity of the solution that has been diluted to 1 liter: 

The volume = 1 litermoles = 0.001


Molarity = moles of solute/ volume in liters    = 0.001moles/1 liter.    
          = 0.001M


The concentration therefore = 0.001M





tysm.#gozmit

Answer 2

Hey mate!

Here's your answer!!

Option (b) 0.001 M is the correct answer.

Explanation :

Molarity of N2CO3 solution is

M1 = (W/GMW) × 1000/V

M1 = (2.65/106) × 1000/125

M1 = 0.1 molar

Now using dilution law Molarity of resultant solution M2 is,

M1V1 = M2V2

M2 = M1V1 / V2

M2 = 0.1 × 10 / 1000

M2 = 0.001 molar

Concentration of resultant solution is 0.001 M.

$hope \: it \: helps \: you.$
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