Equal volumes of ethylene glycol (molar mass = 62) and water (molar mass = 18) are mixed. The depression in freezing point of water is (given `K_(f)` of water `= 1.86 K mol^(-1)` kg and specific gravity of ethylene glycol is 1.11)

Equal volumes of ethylene glycol and water are mixed. The depression in freezing point of water is (K_(f) of water =1.86 and specific gravity of ethylene glycol is 1.11)

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) ).

What mass of ethylene glycol (molar mass = "62.0 g mol"^(-1) ) must be added to 5.50 kg of water to lower the freezing point of water from 0^(@)C to -10.0^(@)C ( k_(f) for water = "1.86 K kg mol"^(-1) ).

45 g of ethylene glycol C_(2)H_(6)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

45 g of ethylene glycol C_(2)H_(4)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

45 g of ethylene glycol C_(2)H_(6)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

45 g of ethylene glycol C_(2)H_(6)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

Ethylene glycol is used as an antifreeze in a cold cliamate Mass of ethylene glycol which should be added to 4 kg for water to prevent it from freezing at -6^(@)C will be ( K_(f) for water = 1.86 K kg mol^(-1) and molar mass of ethylene glycol = 62 g mol^(-1) )