Ncert Exercises & Solutions

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

HINT: Use formula for depression in freezing point.

Explanation:

Step 1:

∆T_f = (273.15 − 271) K = 2.15 K

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 =

\frac{5}{342} mol

= 0.0146 mol

Therefore, molality of solution, m=

\frac{0.0146 mol}{0.095 kg} = 0.1537 mol {kg}^{- 1}

Step 2:

Applying the relation,

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

Number of mole of glucose = \(\frac{amount \ of \ glucose}{molar\ mass\ of\ glucose}\)= \(\frac{5}{180}\) = 0.0278 mol

Therefore, molality of the glucose solution, m=

\frac{0.0278 mol}{0.095 kg}

= 0.2926 mol {kg}^{- 1}

Step 3:

Applying the relation,

∆Tf = K_f × 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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