Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg.

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg.

(Molar mass in g ${mol}^{- 1}$ for ethylene glycol 62, $K_{f}$ of water = 1.86 K kg ${mol}^{- 1}$ ) [2024]

Ans.
(15)

   $Freezing point of pure water T_{f}^{o} = 0 ° C$

   $Freezing point of ethylene glycol solution T_{f} = - 24 ° C$

   $Depression in freezing point, Δ T_{f} = T_{f}^{o} - T_{f} = 0 ° C - (- 24 ° C) = 24 ° C$

   $Value of difference of two temperatures is same in ° C and K, thus$

   $Δ T_{f} = 24 ° C or 24 K$

   $Δ T_{f} = i \times m \times k_{f}$

   $Assuming glycol to be undissociated in water, i = 1$

$24 = 1 \times m \times 1.86$

   $m = \frac{24}{1.86}$

$\frac{W_{glycol}}{M_{glycol} \times W_{water} (in kg)} = \frac{24}{1.86}$

$\frac{W_{glycol}}{62 \times 18.6} = \frac{24}{1.86}$

   $W_{glycol} = 14880 g = 14.88 kg$

Want Us to Email you About Special Offers & Updates?