Question

The elevation in boiling point of aq. cane sugar solution made by mixing 34.2 g of cane sugar in 1 kg of water is (Kb for water is 0.52 K kg mol-¹)

0.52 K

0.052 K

1.04 K

5.2 K​

Answer 1

Answer:

101.04  

o

C

Explanation:

Answer 2

The elevation in the boiling point of aqueous cane sugar solution is (B)0.052 K.

Given,

Mass of cane sugar=34.2 g

Mass of water=1 Kg

Kb(elevation constant)=0.52 K Kg/mol.

To find,

the elevation in boiling point.

Solution:

• Elevation in boiling point of a solution when a solute is added to it is one of the colligative properties.
• The elevation in boiling point of a solution, ΔTb is given by the following expression:
• ΔTb=Kb x m.
• where, Kb-elevation constant(ebullioscopic constant, it only depend on the nature of the solvent), m-molality of the solution.
• Molality of a solution is defined as the number of moles of a solute present in 1 Kg of solvent.

The molar mass of cane sugar is 342 g.

The moles of cane sugar(solute) present will be:

$Moles=\frac{Mass}{Molar mass} \\Moles=\frac{34.2}{342} \\Moles=0.1.$

The molality of the solution will be:

$Molaity=\frac{Moles}{Mass of solvent(Kg)} \\Molality=\frac{0.1}{1} \\Molaity=0.1 molal.$

The elevation in boiling point of the given solution will be:

ΔTb=Kb x m

ΔTb=0.52 x 0.1

ΔTb=0.052 K.

Hence, the elevation in the boiling point is 0.052 K.

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