Question

Each question is and was separated by an asterisk.

*The freezing point of water is 0.00°C at 1 atmosphere.  
If 14.18 grams of cobalt(II) bromide, (218.7 g/mol), are dissolved in 271.6 grams of water ...
The molality of the solution is  m.
The freezing point of the solution is  °C.

*The freezing point of water H₂O is 0.00°C at 1 atmosphere.
How many grams of sodium nitrate (85.00 g/mol), must be dissolved in 249.0 grams of water to reduce the freezing point by 0.350°C ?
g sodium nitrate.

*An aqueous solution of potassium sulfide has a concentration of 0.439 molal.  
The percent by mass of potassium sulfide in the solution is  %.

Answer 1

1)

mass of CoBr2 = 14.18 g

moles of CoBr2 = 14.18 / 218.74 = 0.0648 mol

molality = moles / mass of solvent in kg

             = 0.0648 / 0.2716

molality = 0.2387 m

delta Tf = i x Kf x m

0 - Tf = 3 x 1.86 x 0.2387

Tf = - 1.33 oC

The freezing point of the solution = - 1.33 oC

2)

delta Tf = i x Kf x m

0.350 = 2 x 1.86 x m

m = 0.0941

molality = moles / 0.249

0.0941 = moles / 0.249

moles = 0.0234

mass of NaNO3 = 1.99 g

3)

mass of K2S = 0.439 x 110.26 = 48.4 g

mass % = 48.4 / (1 + 0.0484) ] x 100

             = 2.17 %