You need to make an aqueous solution of 0.150 M sodium fluoride for an experiment in lab, using a 500 mL volumetric flask. How much solid sodium fluoride should you add?

Explanation:

Solution required is .15 M 500 mL sodium fluoride .

molecular weight of sodium fluoride = 42

one mole of sodium fluoride = 42 grams

No of moles contained in the solution of .15M , 500 mL solution

= .15 x 0.5 moles = .075 moles

1 mole = 42 g

.075 moles = .075 x 42 = 3.15 grams .

So mass of sodium fluoride required = 3.15 g .

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.