What happens when all the colors of white light are reflected?
do they bounce back or do they get absorbed?

They get bounced back. Where the light is turned to opposite of that place,there is usually light

Explanation:

Part A: m = 0.02356 mol/kg = 0.02356 mPart B: Xsolute = 4.243×10⁻⁴Part C: % m/m = 0.1376%Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)m = n solute / Kg solventX solute = n solute / N total% m/m = (mass of solute / mass of solution) × 100ppm = (mass of solute / mass of solution) × 1,000,000density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

            m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg  H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

          X solute = n solute / N total

i) n solute =  2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

         % m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

       ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm