2.7 Kg of each of water and acetic acid are mixed. The freezing point of the solution will be $-\mathrm{x}\,^{\circ}\mathrm{C}$. Consider the acetic acid does not dimerise in water, nor dissociates in water. $\mathrm{x}=$ ____ (nearest integer) [Given : Molar mass of water $=18\,\mathrm{g\,mol}^{-1}$, acetic acid $=60\,\mathrm{g\,mol}^{-1}$] $\mathrm{K}_{\mathrm{f}}$ $\mathrm{H}_{2}\mathrm{O}$: $1.86\,\mathrm{K\,kg\,mol}^{-1}$ $\mathrm{K}_{\mathrm{f}}$ acetic acid : $3.90\,\mathrm{K\,kg\,mol}^{-1}$ freezing point : $\mathrm{H}_{2}\mathrm{O}=273\,\mathrm{K}$, acetic acid $=290\,\mathrm{K}$]

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Solutions

2.7 Kg of each of water and acetic acid are mixed. The freezing point of the solution will be

- x^{\circ} C

. Consider the acetic acid does not dimerise in water, nor dissociates in water.

x =

____ (nearest integer)
[Given : Molar mass of water

= 18 g mol^{- 1}

, acetic acid

= 60 g mol^{- 1}

]

K_{f}

H_{2} O

1.86 K kg mol^{- 1}

K_{f}

acetic acid :

3.90 K kg mol^{- 1}

freezing point :

H_{2} O = 273 K

, acetic acid

= 290 K

]

Answer: 31

Solution

By moles, water (

2700/18 = 150 mol

) is present in far greater molar amount than acetic acid (

2700/60 = 45 mol

), so water is treated as the solvent.

Δ T_{f} = K_{f} \times m = 1.86 \times \frac{45 mol (acetic acid)}{2.7 kg (water)} = 1.86 \times 16.67 \approx 3 1^{\circ} C

x = 31

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Integer answer

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