Question

the molar mass of the solute can be determined by measuring which other quantity?

• the vapor pressure of the solution
• the density of the solution
• either the boiling - point elevation or the density of the solution
• either the vapor pressure or the boiling - point elevation of the solution
• either the vapor pressure, boiling - point elevation, or density of the solution

a solution containing 26.9 g of a molecular compound dissolved in 100.0 g of water has a boiling point of 101.5 °c. calculate the molar mass of the compound. the $k_{b}$ for water is 0.512 °c/m.
molar mass: g/mol

Explanation:

Step1: Calculate boiling - point elevation

The boiling point of pure water is $T_{b}^0 = 100^{\circ}C$. The boiling point of the solution is $T_{b}=101.5^{\circ}C$. The boiling - point elevation $\Delta T_{b}=T_{b}-T_{b}^0$.
$\Delta T_{b}=101.5 - 100=1.5^{\circ}C$

Step2: Use the boiling - point elevation formula

The boiling - point elevation formula is $\Delta T_{b}=K_{b}m$, where $K_{b}$ is the boiling - point elevation constant and $m$ is the molality of the solution. We know $K_{b} = 0.512^{\circ}C/m$ and $\Delta T_{b}=1.5^{\circ}C$. We can solve for $m$ (molality).
$m=\frac{\Delta T_{b}}{K_{b}}=\frac{1.5^{\circ}C}{0.512^{\circ}C/m}\approx2.93m$

Step3: Define molality and solve for moles of solute

Molality $m=\frac{n}{m_{solvent}(kg)}$, where $n$ is the number of moles of solute and $m_{solvent}$ is the mass of the solvent in kilograms. The mass of water (solvent) $m_{solvent}=100.0g = 0.1000kg$. Since $m = 2.93m$, we can find $n$.
$n=m\times m_{solvent}=2.93mol/kg\times0.1000kg = 0.293mol$

Step4: Calculate molar mass

The molar mass $M=\frac{m_{solute}}{n}$, where $m_{solute}=26.9g$ and $n = 0.293mol$.
$M=\frac{26.9g}{0.293mol}\approx91.8g/mol$

Answer:

$91.8g/mol$