Question

270 g of glucose is added to 1 kg of water. The boiling point of this solution is

Answer 1

mass of glucose = 270g

mass of water = 1kg

no. of moles of glucose =$\begin{aligned} \dfrac{270}{180} = 1.5\ mol \end{aligned}$

molality of glucose(m) = $\dfrac{1.5mol}{\frac{1kg}{1kg}} = 1.5mol/kg$

From, elevation in boiling point $T_b = ik_bm$

where $k_b = 0.512\ K.kg/mol$

and i (Vant Hoff's factor) = 1, since glucose is covalent

$\therefore T_b =1× 0.512 ×1.5$

$T_b = 0.768K$

$\therefore$ The boiling point of the solution = 100 + 0.768 = 100.768°C