Question

Kindly try complete all parts in Significant answers sensitive

(1a) n part B of the experiment, you increase the density of a saltwater solution by adding more and more salt (NaCl). Initially, 100.0 mL of water equals 100.0 g of water, as water has a density of approx. 1.0 g/mL.
Calculate the percent composition in % by mass [enter the unit as % (m/m)] as mass of NaCl per 100 g of solution after adding 23.51 g of NaCl to 100.0 g of water

(1b) An irregular shaped solid object consisting of an unknown material has a mass of 340.44 g. The object is placed into a graduated cylinder filled with 151 mL of water. The water level rises to 213 mL. Determine the density of the material in g/mL.

(1c)A cube consisting of an unknown material has a mass of 24.447 g and each side has a length of 1.48 cm.
Determine the density of the material in g/mL.

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Answer 1

#1a. Total mass of solution = Mass of NaCl + Mass of water

                                                = 23.51 g + 100.0 g

                                                = 123.51 g

Now,

            % NaCl (m/ m) = (Mass of NaCl / Mass of Solution) x 100

                                    = (23.51 g / 123.51 g) x 100

                                    = 19.035 %

                                    = 19.0 %

#1b. Volume of the object = Increase in volume in cylinder after placing object in it

                                                = 213.0 mL – 151.0 mL

                                                = 62.0 mL

Now,

            Density of object = (Mass / Volume) of the object

                                                = 340.44 g / 62.0 mL

                                                = 5.5 g/ mL

#1c. Given, length of each side of the cube, a = 1.48 cm

Volume of cube = a³ = (1.48 cm³) = 3.241792 cm³

                                    = 3.241792 mL                                 ; [1 cm³ = 1 mL]

Density of cube = Mass / Volume

                                    = 24.447 g / 3.241792 mL

                                    = 7.541 g/ mL