Problem 1: 0.5844 g of sodium chloride (NaCl) was weighed on a balance. This was placed in a 1L volumetric flask, water was added up to the marked line, and the mixture was mixed well. (Formula volume of NaCl: 58.44 g/mol) Give the concentration of this solution.


Molar amount of sodium chloride (NaCl) 0.5844 g：  0.5844 (g) / 58.44 (g/mol) = 0.01 mol

1 L of 0.01 mol of sodium chloride　→　concentration＝ 0.01 (mol) / 1 (L) = 0.01 (mol/L)


Problem 2: We had a 10 mmol/L sodium chloride solution. 100 µL of this was aspirated using a micropipettor, placed in a 100 mL volumetric flask, and mixed well by adding water up to the marked line. Give the concentration of this solution.

Note the milli (m) and micro (µ) conversions.

Molar volume of sodium chloride in the solution aspirated by micropipettor：

　　　10×10-3 (mol/L)×100×10-6 (L) = 1000×10-9 (mol)

concentration：1000×10-9 (mol) / ( 100×10-3 (L) ) = 10×10-3 (mol/L) ( = 10 mmol/L )


Terakhir diubah: Kamis, 19 Oktober 2023, 16:19