How many grams of alcl3 are required to make a 2.25m solution in 30.0 g of water?

We know that, Molality = 

Given: Molality of solution = 2.25 m
Weight of solvent = 30 g = 0.030 kg
Molecular weight of AlCl3 = 133.34 g/mol

Therefore we have, 2.25 = 
∴, weight of solute (g) = 9.00 g

Thus, 9.00 g of AlCl3 is required to make a 2.25m solution in 30.0 g of water

There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.