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Question:
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Answer:

In case of cane sugar:

ΔT_f= (273.15 - 271) K = 2.15 K

Molar mass of sugar (C₁₂H₂₂O₁₁) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol ^{- 1}

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 - 5)g = 95 g of water.

Now, number of moles of cane sugar =5/342 mol

= 0.0146 mol

Therefore, molality of the solution,m =0.0146mol / 0.095kg

= 0.1537 kg mol^{- 1}

Now applying the relation,

ΔT_f = K_f × m

⇒ K_f = ΔT_f / m

⇒ 2.15K / 0.1537 kg mol^-1

= 13.99 K kg mol^-1

Molar of glucose (C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol^{- 1}

5% glucose in water means 5 g of glucose is present in (100 - 5) g = 95 g of water.

∴ Number of moles of glucose = 5/180 mol

= 0.0278 mol

Therefore, molality of the solution, m =0.0278 mol / 0.095 kg

= 0.2926 mol kg^{- 1}

Applying the relation,

ΔTf = Kf × m

= 13.99 K kg mol^{- 1} × 0.2926 mol kg ^{- 1}

= 4.09 K (approximately)

Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K= 269.06 K.

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Question:
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

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Question: A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

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Question:
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.