Home
Class 12
CHEMISTRY
K(f) for water is 1.86 K kg mol^(-1). IF...

`K_(f)` for water is `1.86 K kg mol^(-1)`. IF your automobile radiator holds `1.0 kg` of water, how many grams of ethylene glycol `(C_(2)H_(6)O_(2))` must you add to get the freezing point of the solution lowered to `-2.8^(@)C` ?

Promotional Banner

Similar Questions

Explore conceptually related problems

K_(f) for waer is 1.86 K kg mol^(-1) . If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C_(2)H_(6)O_(2)) must you add to get the freezing point of the solution lowered to -2.8^(@)C ?

K_(f) for waer is 1.86 K kg mol^(-1) . If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C_(2)H_(6)O_(2)) must you add to get the freezing point of the solution lowered to -2.8^(@)C ?

k_f for water is 1.86 k kg mol^(-1) If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C_2H_6O_2) must you add to get the freezing point of the solution lowered to - 2. 8^@C ?

K_f for water is 1.86K kg mol^-1 . If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C_2H_6O_2) must you add to get the freezing point of the solution lowered to 0.8^@C ?

45 g of ethylene glycol (C_(2) H_(6)O_(2)) is mixed with 600 g of water. The freezing point of the solution is (K_(f) for water is 1.86 K kg mol^(-1) )

45 g of ethylene glycol (C_(2) H_(6)O_(2)) is mixed with 600 g of water. The freezing point of the solution is (K_(f) for water is 1.86 K kg mol^(-1) )

45g of ethylene glycol (C_2H_6O_2) is mixed with 600 g of water. Calculate: the freezing point of the solution. (K_f for water=1.86K kg mol^-1) .