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What is the molarity of a sodium hydroxide solution made by combining 2.0 L of 0.60 M NaOH with 495 mL of 3.0 M NaOH?
Assume the volumes of the two solutions to be additive.​

Respuesta :

Answer:

1.1 M

Explanation:

I git it wrong and it gave me the answer

The molarity of the final solution i.e.  sodium hydroxide solution at the time of combining should be 1.1.

Calculation of the molarity:

Since

Molarity (M1) = 0.60 M

Volume ( V1) = 2.0 L

Molarity ( M2 ) = 3.0 M

Volume ( V2) = 495 ml = 0.495 L

We know that

Moles = Molarity * volume in L

Now

Moles of 1st solution = n1 = M1 * V1

So

n1 = 0.60 * 2.0 = 1.2 mol

Now

Moles of 2nd solution = n2 = M2 * V2

So, n2 = 3 * 0.495 = 1.485 mol

Now

Total moles in final solution = n = n1 + n2

So,

n = 1.2 + 1.485 = 2.685 mol

Now

Total volume is

= V = V1 + V2

= 2.0 +0.495

= 2.495 L

Now

Molarity of final solution = n / V

= 2.685 / 2.495

= 1.0761 M

= 1.1

hence, The molarity of the final solution is 1.1.

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