Question

A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. What is the formula of the salt? Show your calculations.

Answer 1

For the answer to the question above, I can't help you directly because I don't have a calculator right now. But I'll show you how to solve this.
use the freezing point depression formula for this one: delta T = i * m * K where K is a constant, m is the molality (mol solute/kg solvent), and i is the van'hoff factor the van hoff factor is the number of ions that your salt dissociates into. Since it's an ALKALI flouride salt, how many ions? k is just a constant, you get it from a table in your textbook somewhere So you have everything to solve for the molality of the solution, once you did that, multiplying it by the mass of water to find the mols of the salt. Take the mass of the salt and divide by this mols to figure out the molar mass, and then compare it with the periodic table to identify the salt.

Mole solute x mass of Water = Mol solute
kg Solvent

then

Mass of solute x 1 = molar mass
mole of solute

Answer 2

Answer: RbF

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f^0-T_f=(0-(-1.4)^0C=1.4^0C` = Depression in freezing point

i= vant hoff factor = 2 (for alkali metal flouride, MF)

`K_f` = freezing point constant =
`1.86^0C/m`

m= molality

`\Delta T_f=i* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

Weight of solvent (water)= 100 g = 0.1 kg

Molar mass of unknown solute = ? g/mol

Mass of unknown solute added = 4 g

`1.4=2* 1.86* (4g)/(M g/mol* 0.1kg)`

`M=106g/mol`

Thus the mass of
`MF` is 106 g/mol

x + 19 = 106

x = 87 g

Thus the atomic mass is near to that of rubidium, thus the formula of the salt is
`RbF`