Question

A technician tares a 100.0 mL volumetric flask at 150.00 g. After adding sodium chloride to the flask it then weighs 158.84 g. Assuming an error of 0.2 mL in the volumetric volume and 0.005 g in the weight, calculate the molar concentration of sodium chloride and its associated standard deviation.

Answer 1

Step-by-step explanation:

Since, it is known that number of moles equal to the mass divided by the molar mass of a substance.

As molar mass of NaCl is 58.44 g/mol. Hence, number of moles of NaCl present in 158.84 g are calculated as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

=
`(158.84 g)/(58.44 g/mol)`

= 2.715 mol

Therefore, there are 2.715 moles present in 100 ml solvent.

Hence, the molar concentration will be as follows.

Molar concentration =
`(2.715 mol)/(100 * 10^(-3))` mol/liter

= 27.15 M

It is given that error in weight is 0.005 g and error in volume is 0.2 ml.

Hence, total volume will be (100ml + 0.2 ml) = 100.2 ml.

Therefore, now molar concentration will be as follows.

Molar concentration =
`(2.715 mol)/(100.2 * 10^(-3))` mol/liter

= 27.09 M

Thus, calculate the deviation as follows.

(27.15 - 27.09) mol/liter

= 0.06 mol/liter

Hence, we can conclude that the molar concentration of sodium chloride 27.15 M and its associated standard deviation is 0.06 mol/liter.