QJEEM12Cha6c8: JEE Main 2012 Chemistry - Class 12 JEEM12Cha6c8 Solution: K f for water is 1.86K kg mol –1 . If your automobile radiator…

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QJEEM12Cha6c8: JEE Main 2012 Chemistry - Class 12 JEEM12Cha6c8 Solution: K f for water is 1.86K kg mol –1 . If your automobile radiator…

K_f for water is 1.86K kg mol^–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C₂H₆O₂) must you add to get the freezing point of the solution lowered to –2.8^oC ?

Multiple Choice Options

A 72 g
B 93 g
C 39 g
D 27 g

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Accepted Answer

Δ T_{f} = i \times K_{f} \times m

Given

Δ T_{f} = 2.8, K_{f} = 1.86 K k g m o l^{- 1} i = 1

(ethylene glygol is a non- electrolyte)

wt. of solvent

= 1

k g;

Let of wt of solute

= x

Mol. wt of ethyllene glycol

= 62

2.8 = 1 \times 1.86 \times \frac{x}{62 \times 1}

or

x = \frac{2.8 \times 62}{1.86} = 93 g m

QJEEM12Cha6c8: JEE Main 2012 Chemistry - Class 12 JEEM12Cha6c8 Solution: K f for water is 1.86K kg mol –1 . If your automobile radiator…

Multiple Choice Options

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