Question

Calculate the molarity of the following solutions: one prepared from 72.0 g of NaOH in a final volume of 740 mL. and another prepared from 82.0 g of NaOH in a final volume of 420 mL of solution. Which one of these solutions has the higher molar concentration? First solution (72.0 g of NaOH in a final volume of 740 mL) Molarity = 0.002 Second solution (82.0 g of NaOH in a final volume of 420 mL) Molarity = The solution prepared from 82.0 g of NaOH in a final volume of 420 mL has a higher concentration. The solution prepared from 72.0 g of NaOH in a final volume of 740 mL has a higher concentration.

Answer 1

Answer-

Given,

First Solution- 72 g NaOH in 740 ml

Second Solution - 82 g of NaOH in 420 ml

Molar Mass of NaOH = molar mass of Na + molar mass of O + molar mass of H = 39.997 g/mol

We know that,

1 ml = 0.001 L

So, 740 ml = 0.74 L

420 ml = 0.42 L

Molar Concentration or Molarity = Moles/ Volume in Liters

Also, Moles = mass/ Molar Mass

So, Molar Concentration or Molarity = mass / (Molar Mass * Volume in Liters)

For Solution First,

Molar Concentration or Molarity = 72 g / (39.997 g/mol * 0.74L) = 2.43 M [Answer]

For Solution Second,

Molar Concentration or Molarity = 82 g / (39.997 g/mol * 0.42 L) = 4.88 M [Answer]

So, the solution prepared from 82.0 g of NaOH in the final volumeof a 420 mL has a higher concentration.