Calculate the freezing point and the boiling point of each of the following aqueous solutions. (Assume complete dissociation. Assume that water freezes at 0.00°C and boils at 100.000°C. 1.86°C Kf = molal 0.51°C K = molal a. 0.056 m MgCl2 T = °C T = °C b. 0.056 m FeCl3 T = °C T = °C

Calculating molality.

The normal boiling point of a certain liquid X is 110.00 °C, but when 106. g of alanine (C3H,NO₂) are dissolved in 900. g of X the solution boils at 111.9 °C instead. Use this information to calculate the molal boiling point elevation constant K, of X. Round your answer to 2 significant digits. K₁=0 °C.kg mol X S

Determine the vapor pressure of a solution prepared by dissolving 16.8 mol of non-volatile aspirin (MM = 180.15 g/mol) in 0.691 kg of CH3CH2OH (MM = 46.07 g/mol) at 25 °C. Pº ethanol 44.0 torr at 25 °C. Give the answer to 3 significant figures. Do not include units. =

At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.68 atm. Suppose a solution is prepared by mixing 88.0 g of benzene and 64.1 g of thiophene (C4H4S). Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal. 0 atm 0x10 X

The freezing point of water is 0.00 °C at 1 atmosphere. A student dissolves 11.60 grams of cobalt(II) sulfate CoSO4,  (155.0 g/mol), in 279.3 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb(C/m) Kf(C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67   Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02   The molality of the solution is  m. The freezing point of the solution is  °C.

Predict the vapor pressure of a solution prepared by mixing 45.0 g solid Mg3(PO4)2 (molar mass = 262.9g/mol) with 185 g water at 25°C. (Vapor pressure of pure water at 25 C is 23.76 torr.) A 20.1 torr B 20.6 torr C 19.8 torr D 21.6 torr E 21.9 torr

The freezing point of water is 0.00 °C at 1 atmosphere. How many grams of cobalt(II) acetate (177.0 g/mol), must be dissolved in 245.0 grams of water to reduce the freezing point by 0.300 °C? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula Mass Water H₂O Ethanol CH3 CH₂OH 1.22 Chloroform CHC13 3.67 Benzene C6H6 2.53 Diethyl ether CH3 CH₂ OCH2 CH3 2.02 Kb (°C/m) Kf(°C/m) 0.512 g 1.86 1.99 5.12

dentify which ONE of the following aqueous solutions has the highest freezing point assuming 100% dissolution.   a. 2.00 m KF     b. 1.75 m MgBr2     c. 1.50 m Al2(SO4)3     d. 1.25 m (NH4)3PO4     e. 5.00 m glucose (C6H12O6)

The freezing point of water is 0.00 °C at 1 atmosphere. A student dissolves 11.60 grams of cobalt(II) sulfate CoSO4,  (155.0 g/mol), in 279.3 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb(C/m) Kf(C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67   Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02   The molality of the solution is  m. The freezing point of the solution is  °C.