Asalt 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.

RbF

Explanation:

Depression in freezing point is given by:

= Depression in freezing point

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

= freezing point constant =

m= molality

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

Molar mass of unknown solute = ? g/mol

Mass of unknown solute added = 4 g

Thus the mass of 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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :