Cacl_2 is often used to melt ice on sidewalks.
a) could cacl_2 melt ice at -33°c? yes or no? assume that the solubility of cacl_2 at this temperature is 70.1 g caci_2/100.0 g of h2o and that the van't hoff factor for a saturated solution of cacl_2 is i = 2.5.
b) what is the minimum amount of cacl_2 that would be required to melt ice on sidewalks at the above temperature? use 100.0 g of water as the amount of solvent.

a) No, it couldn't

b) 78.81 g

Explanation:

When a nonvolatile solute is added to a pure solvent, the melting point of the solvent must decrease, which is called cryoscopy. The temperature variation (melting point of pure solvent - the melting point of solution) can be calculated by:

ΔT = Kc.W.i

Where Kc is the cryoscopy constant (for water it's 1.86 °C/mol.kg), W is the molality, and i is the van't Hoff factor of the solute.

W =m1/m2*M1

Where m1 is the mass of the solute (in g), m2 is the mass of the solvent (in kg), and M1 is the molar mass of the solute (CaCl2 = 111.0 g/mol). The van't Hoff factor indicates how much of the solute ionizes in the solution.

a) The melting point of water is 0°C, so let's calculate the new melting point with the given information:

m1 = 70.1 g

m2 = 100 g = 0.1 kg

W = 70.1/(0.1*111) = 6.31 mol/kg

ΔT = 1.86*6.31*2.5

ΔT = 29.34°C

0 - T = 29.34

T = -29.34°C

Thus, at -33°C, the ice will not melt yet.

b) Let's then found out the value of m1 in this case:

ΔT = Kc.W.i

0 - (-33) = 1.86*W*2.5

4.65W = 33

W = 7.1 mol/kg

W = m1/M*m2

7.1 = m1/(111*0.1)

m1 = 78.81 g