Atechnician 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.

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 =

                                =

                                 = 2.715 mol

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

Hence, the molar concentration will be as follows.

       Molar concentration = 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 = 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.

explanation:

i think option no d is the correct answer

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g