Question

if a 359 ml glass of water (1 g/ml) can contains 0.962 moles of sodium bicarbonate, what is the boiling point (°c)? kb 0.512 °c/m (enter two digits after the decimal)

Explanation:

Step1: Calculate the molality of the solution

First, find the mass of water. Given volume of water $V = 359\ mL$ and density $
ho=1\ g/mL$, so mass of water $m_{water}=
ho V=359\ g = 0.359\ kg$. Molality $m=\frac{n}{m_{water}}$, where $n = 0.962\ mol$ (moles of sodium bicarbonate). So $m=\frac{0.962\ mol}{0.359\ kg}\approx2.68\ m$.

Step2: Consider the van - 't Hoff factor

Sodium bicarbonate ($NaHCO_3$) dissociates into $Na^+$ and $HCO_3^-$ in water, so the van - 't Hoff factor $i = 2$.

Step3: Calculate the boiling - point elevation

The formula for boiling - point elevation is $\Delta T_b=iK_bm$. Given $K_b = 0.512^{\circ}C/m$, $i = 2$, and $m\approx2.68\ m$. Then $\Delta T_b=2\times0.512^{\circ}C/m\times2.68\ m\approx2.75^{\circ}C$.

Step4: Find the boiling point of the solution

The boiling point of pure water is $T_{b0}=100^{\circ}C$. The boiling point of the solution $T_b=T_{b0}+\Delta T_b = 100^{\circ}C + 2.75^{\circ}C=102.75^{\circ}C$.

Answer:

$102.75$