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JEE Main levelSingle correct optionDifficulty 2 of 3so194

A $10%$ by mass aqueous solution of glucose (molar mass $180 g mo l^{- 1}$ ) is heated at $1 atm$ . Taking $K_{b} = 0.52 K kg mo l^{- 1}$ , the boiling point of the solution is closest to

A100.05 $^{\circ} C$
B100.29 $^{\circ} C$
C100.64 $^{\circ} C$
D100.32 $^{\circ} C$

Answer: D

In

100 g

of solution there are

10 g

glucose (

0.0556 mol

) and

90 g

water. So

m = 0.0556/0.090 = 0.617 mol k g^{- 1}

and

Δ T_{b} = 0.52 \times 0.617 = 0.321 K

. The boiling point is about

100.3 2^{\circ} C

.

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