Equal volumes of ethylene glycol (molar mass = 62) and water (molar mass =18 ) are mixed. The depression in freezing point of water is (given K f of water = 1.86 K mor -1 kg and specific gravity of ethylene glycol is 1.11 )

Question

Equal volumes of ethylene glycol (molar mass = 62) and water (molar mass =18 ) are mixed. The depression in freezing point of water is (given K f of water = 1.86 K mor -1 kg and specific gravity of ethylene glycol is 1.11 ) (A) 0.0033 (B) 0.033 (C) 0.33 (D) 33 .3

Tardigrade Admin · Accepted Answer

M=D × V M (Ethylene glycol) =1.11 × v M (water) =1.00 × v n (Ethylene glycol) =(1.11 v/62) mol Δ Tf=Kf × m=1.86 × (1.11 v/62) × (1000/1.00 v) =(1.86 × 1.11 × 1000/62)=33.3

Q. Equal volumes of ethylene glycol (molar mass $= 62$ ) and water (molar mass $= 18$ ) are mixed. The depression in freezing point of water is (given $K_{f}$ of water $=$ $1.86 K$ mor $^{- 1} k g$ and specific gravity of ethylene glycol is $1.11$ )

0.0033

0.033

0.33

33 .3

$M = D \times V$
$M$ (Ethylene glycol) $= 1.11 \times v$
$M$ (water) $= 1.00 \times v$
$n$ (Ethylene glycol) $= \frac{1.11 v}{62} m o l$
$Δ T_{f} = K_{f} \times m = 1.86 \times \frac{1.11 v}{62} \times \frac{1000}{1.00 v}$
$= \frac{1.86 \times 1.11 \times 1000}{62} = 33.3$

Q. Equal volumes of ethylene glycol (molar mass =62) and water (molar mass =18 ) are mixed. The depression in freezing point of water is (given Kf​ of water = 1.86K mor −1kg and specific gravity of ethylene glycol is 1.11 )