Question

A. The boiling point of water is 100.0°C at 1 atmosphere.

How many grams of chromium(II) chloride (122.9 g/mol), must be dissolved in 256.0 grams of water to raise the boiling point by 0.300°C? Refer to the table for the necessary boiling or freezing point constant.

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

____g chromium(II) chloride.

B. The boiling point of water is 100.00 °C at 1 atmosphere.

A student dissolves 10.44 grams of nickel(II) nitrate, Ni(NO₃)₂ (182.7 g/mol), in 291.0 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 boiling point of the solution is ____°C.

Answer 1

A)

Let; We have given;

Mass of Solvent(water) =256.0 g =0.256Kg  

Boiling point of solvent(Water) = 100^oC

Kb for water = 0.512 ^oC/m

Elevation in Boiling Point (T_b) = 0.300^oC

We know;

T_b = i x Kb x m Where; i =Vant Hoff factor For; CrCl₂ = 3 as it forms three ions 1Cr²⁺ + 2 Cl^-

Putting the known values in the above equation we get;

0.300^oC = 3 x 0.512 ^oC/m x molality(m)

Molality(m) = 0.1953m of CrCl₂ solution

Now; We know; Molality = Number of moles / Kg of solvent

01953 m =Number of moles /0.256 Kg

Number of moles of CrCl₂ = 0.05 moles

Converting these moles to grams by using the given molar mass of CrCl₂;

0.05 moles of CrCl₂ x ( 122.9 g/ 1mole) = 6.145 g of CrCl₂

_B)

Let; We have given;

Mass of Solvent(water) =291.0 g =0.291Kg  

Boiling point of solvent(Water) = 100^oC

Kb for water = 0.512 ^oC/m

Mass of solute [Ni(NO₃)₂] =10.44 g

Molar mass of solute [Ni(NO₃)₂] = 182.7 g/mol

Calculating the moles of Ni(NO₃)₂; 10.44g x( 1mole/ 182.7g) =0.05714 moles of [Ni(NO₃)₂]

Now; Molality = 0.05714 moles/0.291 Kg

Molality= 0.1964 m

We know;

T_b = i x Kb x m Where; i =Vant Hoff factor For; Ni(NO₃)₂ = 3 as it forms three ions 1NI²⁺ + 2 NO₃^-

Putting the known values in the above equation we get;

T_b= 3 x 0.512 ^oC/m x 0.1964m

T_b = 0.3016 ^oC

Now; We know; T_b = T_solution - T_solvent

  0.3016 ^oC = T_solution - 100^oC

T_solution = 100.3^oC