The freezing point of pure water is 0.0°C. How many grams of ethylene glycol (C2H6O2) must be mixed in 100.0 g of water to lower the freezing point of the solution to -8.8°C? g

17.54 g

Explanation:

the freezing point of the solution to -8.8°C

depression in freezing point = 8.8°C

Molal freezing point depression constant of ethylene glycol  Kf = 3.11⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

Let gram of ethylene glycol required be x .

moles = m / mol weight of ethylene glycol

= x / 62

100 g of water = .1 kg

moles of solute dissolved in 1 kg of water

m = x / 62 x .1

m = 10 x / 62

Using the above equation for depression in freezing point

8.8 = 3.11 x 10 x / 62

x = 17.54 g .

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.