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250 mL of a sodium carbonate solution contains 2.65 g of Na2CO3 10 mL of this solution is added to x mL of water to obtain 0.001 M Na2CO3 solution. What is the value of x in mL? (Molecular Mass of Na2CO3 106)

250 mL of a sodium carbonate solution
 
contains 2.65 g of Na2CO3 10 mL of this
solution is added to x mL of water to obtain
0.001 M Na2CO3 solution. What is the
value of x in mL?
(Molecular Mass of Na2CO3 106)

Grade:12

2 Answers

Arun
25750 Points
4 years ago
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
 
 
Khimraj
3007 Points
4 years ago
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

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