Question

A radiator was filled with 10L of water to which 2.5L of methanol ( density = 0.8 gm/L) were added. At 9:00 pm, the vehicle is parked outdoors where the temperature is $0{}^\circ C$. The temperature is decreasing at a uniform rate of $0.5{}^\circ C/\min $. Upto what time will there be no danger to the radiator of the car. ${{K}_{f}}(water) = 1.86Kg.mo{{l}^{-1}}$. Assume methanol to be non-volatile.
A. 11.67 min
B. 23.25 min
C. 32.56 min
D. 65.1 min

Answer 1

Hint: Depression of freezing point is lowering of temperature of solvents when a solute is added. It is a colligative property which is directly proportional to the molality of the solution. It can be described by the formula:
$\Delta {{T}_{f}} = i\times {{K}_{f}}\times m$
 $\Delta {{T}_{f}}$ = Freezing point depression
$i$= Vant Hoff factor
${{K}_{f}}$ = cryoscopic constant
m = molality

Complete step by step solution:
We have given the following things
Volume of water= 10L
2.5 L of Methanol
Temperature = $0.5{}^\circ C/\min $
${{K}_{f}}(water) = 1.86kg.mo{{l}^{-1}}K.$

Now, Density of Methanol is 0.8gm/ml
Hence, Mass of 2.5 L of Methanol = 0.8 × 2500 ml = 2000 g
The molar mass of methanol is $\dfrac{2000}{32}$ = 62.5 moles
$D = \dfrac{Mass}{Volume}$
Volume of water is 10 L
Density of water is 1Kg/L
Hence, the mass of water will be 10 kg.
Molality of methanol will be = $\dfrac{62.5}{10}$ = 6.25 m

The depression in the freezing point will be
$\Delta {{T}_{f}} = i\times {{K}_{f}}\times m$
$\Delta {{T}_{f}}$ = 1.86 × 6.25 = $11.62{}^\circ C$
Rate of decrease in temperature = $0.5{}^\circ C/\min $
Time for which there will be no danger to the radiator of the car = $\dfrac{11.62}{0.5} = 23.25\min $
So, the correct answer is “Option B”.

Note: Van't Hoff Factor is the ratio between the number of moles of particles formed in solution per mole of solute. Since Methanol is a non-electrolyte and it does not dissociate into ions therefore, its Van't Hoff Factor is 1.