Question

Can somebody solve this question for me? ​

Answer 1

Given,

$k_{b}$ of the water: 0.52 K kg $mol^{-1}$

Amount of glucose: 18g

Pressure: 1.013 bar

Water mixed: 1kg

To Find,

The temperature at which the water will boil.

Solution,

ΔT = $K_{b}$ × ( $W_{B}$× 1000)$__$

                  ________

                  ( $m_{B}$×$W_{A}$)

ΔT = 0.52 × (18 × 1000)/ (180 × 1000)

T - $T_{0}$ =  0.052

T = 0.052 + 373

T = 373.052 K

T is the temperature that is required to be found.

The boiling point of pure water is 373K.

By cross multiplying we can solve it with terms to T and find the temperature in kelvin unit.

Hence, the temperature at which the water will boil after the glucose is mixed is at 373. 052k.