CBSE Class 11-science Answered
250 ml of Na2CO3 sol. contains 2.65 g of Na2CO3. If 10 ml of this solution is diluted to one litre what is the concentration of the resultant solution?
In a gas S and O are 50% by mass,their mole ratio is ____.
Asked by vijaygireesh | 04 May, 2019, 07:42: PM
Expert Answer
Moles of 2.65 g of Na₂CO₃ (molar mass = 106)
Moles = mass/molar mass
= 2.65 g / 106
Moles = mass/molar mass
= 2.65 g / 106
= 0.025 moles
Molarity of this solution=Moles odd solute/Volume in litre
Molarity = 0.025 moles / 0.25 liters (250ml in liters)
Molarity of this solution=Moles odd solute/Volume in litre
Molarity = 0.025 moles / 0.25 liters (250ml in liters)
= 0.1 M
Now using dilution law Molarity of resultant solution (M2) is
Now using dilution law Molarity of resultant solution (M2) is
M1V1 = M2V2
M2 = M1V1 / V2M2
= (0.1 ×10) / 1000
= 0.001
So molar Concentration of resultant solution is 0.001M
in second question,
Dividing percentage by thier atomic mass
S : O
50\32: 50\16
1 : 2
So Mole ratio will be 1:2
Answered by Ravi | 05 May, 2019, 04:35: PM
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