Abeaker contains 450 g of water (h2o). if 9.2 g of calcium fluoride (caf2) is added, what is the molality concentration of the solution?

Molality = number of mole of solute in  1kg= 1000g solvent.

Solute is CaF2.
Molar mass (CaF2) = M(Ca) +2M(F) = 40.1 + 2*19.0 = 78.1 g/mol

9.2 g CaF2 * 1 mol CaF2/78.1 g CaF2 = 0.12 mol CaF2

Solvent is water 450 g.

0.12 mol CaCl2 450 g water
x mol CaCl2    1000 g water

x=0.12*1000/450 = 0.27 

Molality is 0.27 (mol CaF2 in 1 kg water).

Outermost (valence) electrons

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts