Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.45 m AgNO3(aq). Constants may be found here.

Answer 1

Answer: According to the question , our solution is in water so by using the formula ,

delta t = m x Kf x i

we have 2 particles when AgNO3 dissolves so i = 2

Hence, delta t = 2.45 * 1.86 * 2 = 9.114 ⁰C

Hecen the freezing point is comes out to be - 9.114 ⁰C .

now , For the boiling point we have to use ,

delta t = m x Kb x I

delta t = 2.45 * 0.512 * 2 = 2.5088 ⁰C

Hecne the solution is boil at 102.5088 ⁰C

This is all about the given question . Thank you :)