Answer:
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
Calcium chloride cannot melt the ice at -33°C. Cryoscopy is the method of determining a decrease in melting point due to dissolved substances.
The decrease in the melting point of a pure solvent when a non-volatile solute is added to the solution.
[tex]\Delta T = K_c. W.i[/tex]
Where
[tex]K_c[/tex] - cryoscopy constant= 1.86 °C/mol.kg for water
[tex]W[/tex] - molality = 6.31 mol/kg
[tex]i[/tex] - Van't Hoff factor of the solute = 2.5
Put the values in the formula,
[tex]\Delta T = 1.86\times 6.31\times 2.5\\\\\Delta T = \rm \ 29.34^oC[/tex]
Therefore, Calcium chloride cannot melt the ice at -33°C.
Learn more about cryoscopy:
https://brainly.com/question/21854669
