Question

The freezing point of water, H₂O, is 0.000 °C at 1 atmosphere. K_f(water) = -1.86 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 11.70 grams of the compound were dissolved in 205.9 grams of water, the solution began to freeze at -0.587 °C. The compound was also found to be nonvolatile and a non-electrolyte.  

What is the molecular weight they determined for this compound ?  

________ g/mol

Answer 1

Given:-

K_f of water(H₂O) = - 1.86 ⁰C / m

weight of water(H₂O) = 205.9 g

temperature (T₁) = 0.0 ⁰C

temperature (T₂) = - 0.587 ⁰C

weight of compound = 11.70 g

molecular weight of compound =?

As we know that

depression in freezing point (T_f) = temprature of pure solvent - temprature of solution

depression in freezing point (T_f) = temprature of pure water - temprature of solution

depression in freezing point (T_f) = 0.000 - (- 0.587)

depression in freezing point (T_f) =  0.587 ⁰C

As we know that

depression in freezing point (T_f) = K_f m

where m = molality of the solution

K_f = freezing point constant of water

therefore above equation can be written as

depression in freezing point (T_f) = K_f W_compound / M_compound1000/ W_water

where

W_compound = weight of compound

M_compound = molecular weight or mass of compound

W_water =  weight of water

T_f = K_f W_compound / M_compound1000/ W_water

M_compound = 1000 W_compound K_f / W_waterT_f

M_compound = 1000 11.70 1.86 / 205.9 0.587

M_compound = 21762 / 120.8633

M_compound = 180.054 g / mol ( i.e the answer)