Question

find the boiling point of a solution containing 0.520 g of glucose dissolved in 80.2 g of water. given
k for water 1.86 K per mole.

Answer 1

answer for this solution is this, it is 100.066 it is the answer

Answer 2

Answer : The boiling point of a solution is, 373.066 K

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

$T_b-T^o_b=\frac{1000\times k_b\times w_2}{w_1\times M_2}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of pure water = $100^oC=273+100=373K$

$k_b$ = boiling point constant  = 1.86 K/mole

m = molality

$w_2$ = mass of solute (glucose) = 0.52 g

$w_1$ = mass of solvent (water) = 80.2 g

$M_2$ = molar mass of solute (glucose) = 180 g/mole

Now put all the given values in the above formula, we get the boiling point of a solution.

$T_b-373K=\frac{1000\times 1.86K/mole\times 0.52g}{80.2g\times 180g/mole}$

$T_b=373.066K$

Therefore, the boiling point of a solution is, 373.066 K