50 cm3 ethylene glycol is dissolved in 50g water. Freezing point of solution is -34° C Assuming ideal behaviour than density of ethylene glycol is (for water, Kf=1.86 K kg mol-1 )

Question

50 cm3 ethylene glycol is dissolved in 50g water. Freezing point of solution is -34° C Assuming ideal behaviour than density of ethylene glycol is (for water, Kf=1.86 K kg mol-1 ) (A)  1.273 text g/cm3  (B)  2.173 text g/cm3  (C)  1.133 text g/cm3  (D)  2.612 text g/cm3

Tardigrade Admin · Accepted Answer

Volume of ethylene glycol =50 text cm3 ∴ Mass of ethylene glycol (w)=50× d And, W=50g Δ T=0-(-34)=34 Kf=1.86K log mol-1 ∴ Δ Tf=(1000× kf× w/62× 50) (Mol. mass of ethylene glycol= 62) or d=(34× 62× 50/1000× 1.86× 50) or d=1.33 g/cm3

Q. $50 c m^{3}$ ethylene glycol is dissolved in 50g water. Freezing point of solution is $- 34^{\circ} C$ Assuming ideal behaviour than density of ethylene glycol is (for water, $K_{f} = 1.86 K k g m o l^{- 1}$ )

$1.273 g / c m^{3}$

$2.173 g / c m^{3}$

$1.133 g / c m^{3}$

$2.612 g / c m^{3}$

Q. 50cm3 ethylene glycol is dissolved in 50g water. Freezing point of solution is −34∘C Assuming ideal behaviour than density of ethylene glycol is (for water, Kf​=1.86Kkgmol−1 )

1.273g/cm3

2.173g/cm3

1.133g/cm3

2.612g/cm3

Volume of ethylene glycol =50cm3 ∴ Mass of ethylene glycol (w)=50×d And, W=50g ΔT=0−(−34)=34 Kf​=1.86Klogmol−1 ∴ ΔTf​=62×501000×kf​×w​ (Mol. mass of ethylene glycol= 62) or d=1000×1.86×5034×62×50​ or d=1.33g/cm3

Q. $50 c m^{3}$ ethylene glycol is dissolved in 50g water. Freezing point of solution is $- 34^{\circ} C$ Assuming ideal behaviour than density of ethylene glycol is (for water, $K_{f} = 1.86 K k g m o l^{- 1}$ )

$1.273 g / c m^{3}$

$2.173 g / c m^{3}$

$1.133 g / c m^{3}$

$2.612 g / c m^{3}$