10 gram of `C_4H_10` is mixed with 200 gram of `C_6H_6` (l) , then calculate freezing point of solution. [Given for `C_6H_6 K_F = 5.12K Kg/mole & freezing point = 5.5^@C`]

1.8g of fructose (C_(6)H_(12)O_(6)) is added to 2 kg of water. The freezing point of the solution is

Freezing Point Depression: when 15.0 grams of ethyl alcohol (C_(2)H_(5)OH) is dissolved in 750 grams of formic acid, the freezing point of the solution is 7.20^(@)C . The freezing point of pure formic acid is 8.40^(@)C Evalulate K_(f) for formic acid. Strategy: First calculate the molality and the depression of the freezing point. Then solve the equation DeltaT_(f)=K_(f)m for K_(f) by substituting values for m and DeltaT_(f) .

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. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

The freezing point of a solution of 2.40 g of biphenyl( C_(12)H_(10) ) in 75.0 g of benzene ( C_(6)H_(6) ) is 4.40^(@)C . The normal freezing point of benzene is 5.50^(@)C . What is the molal freezing point constant (^(@)C//m) for benzene ?

C_6H_6 freezes at 5.5^(@)C . The temperature at which a solution 10 g of C_4H_(10) in 200g of C_6H_6 freeze is __________ ""^(@)C . (The molal freezing point depression constant of C_6H_6 is 5.12^(@)C//m .)