If a label on the back of a a can of soup indicates that there are 667 mg of sod

Question

If a label on the back of a a can of soup indicates that there are 667 mg of sodium per 1/2 cup, what is the salt concentration of the soup in % (w/v) with w in grams and v in milliters? Only allowed to use the conversions in the photo. Not sure how to.

For each compound, indicate whether the resistance of an aqueous solution of the compound would be high or low.

Na2SO4

C6H12O6

KCl

CH4N2O

Unit Conversions between English and Metric Systems 1 inch = 2.54 cm 1 mile = 1.609 km 1 m = 39.37 inch 1 ft = 12 in 1 qt = 0.946 L 1 L = 33.81 fl oz 1 fl oz = 29.57 mL 1 qt = 4 cup 1 gal = 4 qt 1 kg = 2.205 lb 1 lb = 453.6 g 1 oz = 28.35 g

Explanation / Answer

in order to calculate % (w/v) of a solution, it means w gram of solid is dissolved in v ml of solvent (for percent calculation we have to calculate in 100 ml).

so 667 mg = 0.667 g

now, 4cup = 1qt =0.946 L = 946 mL

0.5 cup = 946*0.5/4 = 118.25 mL

so, 118.2 mL of solution contains 0.667 g of salt

100 ml of solution contains 0.667* 100/118.2

= 0.56 %

so the salt concentration of the soup is 0.56 %

The resistance of a solution is the measure of the electrical conductivity of the solution.

So for the first solution contains Na2SO4 will have high resistance in aqueous solution, because it is a salt and in solution, it completely dissociated in their ions, ions causes high electrical conductivity

for C6H12O6 it will have very less conductivity, Because of the nonionic nature

KCl again it is a salt, highly dissociated in solution so have higher conductance hence have higher resistance

Ch4N2O, urea has little conductance in solution.

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