Question

250 ml of a sodium carbonate solution contains 2.65 grams of 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)

Answer 1

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 liters
Then ;   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 liter
moles = 0.001

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

The concentration therefore = 0.001M

Answer 2

Molarity of N2CO3 solution is
M1 = (W/GMW) * 1000/V
= (2.65/106) * 1000/250
= 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 0.001M