Question

1. how many grams of fuctose (c6h12o6) must be dissolved in 419 g of benzene to raise the boiling point by 3.8 °c? (normal boiling point = 80.1 °c and kb = 2.67 °c/m)

Explanation:

Step1: Calculate the molality (m)

The boiling - point elevation formula is $\Delta T_b = K_bm$. We know $\Delta T_b=3.8^{\circ}C$ and $K_b = 2.67^{\circ}C/m$. Rearranging for $m$, we get $m=\frac{\Delta T_b}{K_b}$.
$m=\frac{3.8^{\circ}C}{2.67^{\circ}C/m}\approx1.423m$

Step2: Calculate the moles of fructose

Molality ($m$) is defined as moles of solute per kilogram of solvent. The mass of benzene (solvent) is $419g = 0.419kg$. Using the formula $m=\frac{n}{m_{solvent}(kg)}$, we can find the number of moles ($n$) of fructose.
$n = m\times m_{solvent}(kg)$
$n=1.423m\times0.419kg\approx0.596mol$

Step3: Calculate the molar mass of fructose

The molar mass of $C_6H_{12}O_6$:
$M=(6\times12.01g/mol)+(12\times1.01g/mol)+(6\times16.00g/mol)$
$M = 72.06g/mol+12.12g/mol + 96.00g/mol=180.18g/mol$

Step4: Calculate the mass of fructose

Using the formula $m = n\times M$, where $n$ is the number of moles and $M$ is the molar mass.
$m=0.596mol\times180.18g/mol\approx107.4g$

Answer:

107.4g