Question

Calculate the boiling point of a solution containing 0.36g of glucose (C6H12O6) dissolved in 100g of water. (given :kb= 052km-1​

Answer 1

Answer:

see the attachment given

Answer 2

The boiling point of a solution containing 0.36g of glucose (C6H12O6) dissolved in 100g of water is 100.014 °c.

Given:

0.36g of glucose (C6H12O6) dissolved in 100g of water.

To Find:

The boiling point of a solution containing 0.36g of glucose (C6H12O6) dissolved in 100g of water.

Solution:

To find the boiling point we will follow the following steps:

As we know,

Elevation in boiling point (Tf - Ti) = kbm

Here, Tf is the temperature after adding solute and Ti is the temperature before which is 100° c.

m is the molarity and kb is constant.

We have to calculate Tf because adding non-solute elevated the temperature of the solution.

The molecular mass of glucose is 180gram.

molarity =

$\frac{given \: mass}{molecular \: mass \times mass \:solventt \: in \: gram \: } \times 1000 = \frac{0.36}{180 \times 100} \times 1000 = 0.02\: m$

Tf - Ti = 0.52 × 0.02 =0.014 °C.

Ti = 100°C

Tf = Ti + 0.014 = 100+0.014 =100.014°C

Henceforth, the boiling point of a solution containing 0.36g of glucose (C6H12O6) dissolved in 100g of water is 100.014°C.