Answer:
-30 °C
Explanation:
First, we have to calculate the molality (m) of the solution. If the solution is 50% C₂H₆O₂ by mass. It means that in 100 g of solution, the are 50 g of solute (C₂H₆O₂) and 50 g of solvent (water).
The molar mass of C₂H₆O₂ is 62.07 g/mol. The moles of solute are:
50 g × (1 mol / 62.07 g) = 0.81 mol
The mass of the solvent is 50 g = 0.050 kg.
The molality is:
m = 0.81 mol / 0.050 kg = 16 m
The freezing-point depression (ΔT) can be calculated using the following expression.
ΔT = Kf × m = (1.86 °C/m) × 16 m = 30 °C
where,
Kf: freezing-point constant
The normal freezing point for water is 0°C. The freezing point of the radiator fluid is:
0°C - 30°C = -30 °C
F= Force
M=Mass
A= acceleration
F=N
Mass= in grams or kilo grams (mostly kg)
A= m/s
Answer:
Icy roads
Explanation:
There is so little friction you slide on it way more than other roads. :)
