How many grams of ethylene glycol (molar mass = 62) should be added to 10 kg of water, so that the resulting solution freezes at `-10^(@)C` (`K_(f)` for water = 1.86 K `"mol^(-1)`).

Explanation

`W_B=?, M_B = 62 g mol^(-1), W_A = 10 kg, K_f = 1.86 k kg mol^(-1)`.
`Delta T_f = 0 - (-10^@C)= 10^@C`
`Delta T_f = 10 K`
`Delta T_f = K_f xx (W_B)/(M_B)xx (1)/(W_A(kg.))`
`W_B =(Delta T_f xx M_B xx W_A (kg.))/(K_f)`
`=(10K xx 62 g mol^(-1) xx 10 kg)/(1.86 K Kg.mol^(-1))= 3.33 kg.`