The freezing point of a solution containing 50 cm 3 of ethylene glycol in 50 g of water is found to be -34- C- Assuming ideal behaviour- calculate the density of ethylene glycol- K f - for water -1-86 K kg mol -1 A- 2-00 g - mLB- 1-13 g - mLC- 2-96 g - mLD- 4-32 g - mL

The correct option is B 1.13 g/mLΔ T = K_f× W_2×1000/Mw_2× W_1 34= 1.86×W_2× 1000/62×50 W_2=56.66 g Volume = W_2/d Density = 56.66/50 = 1.13 g cm^ 3 ...

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Byju's Answer

Standard XII

Chemistry

Depression in Freezing Point

The freezing ...

Question

The freezing point of a solution containing $50 {cm}^{3}$ of ethylene glycol in $50 g$ of water is found to be $- 34^{\circ} C$ . Assuming ideal behaviour, calculate the density of ethylene glycol. ( $K_{f}$ for water $= 1.86 {K kg mol}^{- 1}$ )

2.00 g/mL

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1.13 g/mL

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2.96 g/mL

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4.32 g/mL

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Solution

The correct option is B $1.13 g/mL$
$Δ T = \frac{K_{f} \times W_{2} \times 1000}{M w_{2} \times W_{1}}$
$34 = 1.86 \times \frac{W_{2} \times 1000}{62 \times 50}$
$W_{2} = 56.66 g$
$V o l u m e = \frac{W_{2}}{d}$
$D e n s i t y = \frac{56.66}{50}$
$= 1.13 g c m^{- 3}$

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The freezing point of a solution containing 50 cm3 of ethylene glycol in 50 g of water is found to be −34∘ C. Assuming ideal behaviour, calculate the density of ethylene glycol. (Kf for water =1.86 K kg mol−1)

The correct option is B 1.13 g/mLΔT=Kf×W2×1000Mw2×W1 34=1.86×W2×100062×50 W2=56.66 g Volume=W2d Density=56.6650 =1.13 g cm−3

The freezing point of a solution containing $50 {cm}^{3}$ of ethylene glycol in $50 g$ of water is found to be $- 34^{\circ} C$ . Assuming ideal behaviour, calculate the density of ethylene glycol. ( $K_{f}$ for water $= 1.86 {K kg mol}^{- 1}$ )

The correct option is B $1.13 g/mL$
$Δ T = \frac{K_{f} \times W_{2} \times 1000}{M w_{2} \times W_{1}}$
$34 = 1.86 \times \frac{W_{2} \times 1000}{62 \times 50}$
$W_{2} = 56.66 g$
$V o l u m e = \frac{W_{2}}{d}$
$D e n s i t y = \frac{56.66}{50}$
$= 1.13 g c m^{- 3}$