250 ml of sodiumcarbonate solutioncontains 2.65 g ofNa2CO3.if 10 ml of

1.	250 ml of sodiumcarbonate solutioncontains 2.65 g ofNa2CO3.if 10 ml of this solution isdiluted to500ml the concentrationof the diluted acid will bea) 0.01 Mb)0.001Mc)0.05Md)0.002M
Answer» 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 ofNa₂CO₃ in 10ml solution: If molarity = moles of solute/ volume in litersThen ; Moles = volume in liters x Molarity Moles ofNa₂CO₃ 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

250 ml of sodiumcarbonate solutioncontains 2.65 g ofNa2CO3.if 10 ml of this solution isdiluted to500ml the concentrationof the diluted acid will bea) 0.01 Mb)0.001Mc)0.05Md)0.002M

Answer»

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 ofNa₂CO₃ in 10ml solution:

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

Moles ofNa₂CO₃ 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