Question

COLLIGATIVE PROPERTIES- HOMEWORK 1 Calculate the bp o ate the bp of a solution made by dissolving 20.0g of quinone (CH.03) acetone (bp of acetone- 56 5C, k, = 1 67'Cimolal). 2A sample of comp of compound X weighing 1 2869 was dissolved in 259 benzone K5 12S pe 5.50°C) and the fo of the resulting solution was found to be 3.27 Suiting solution was found to be 3 27°C What is the MW of X7 og of a compound (MW-92) are dissolved in 100g of water, what if the bp of the so Given K 0.512"C/molal 4. When 40.09 when 40. Og of a compound Y are dissolved in 500g benzene (K-2.53°C molat bpou the resulting solution has a bpe 81.4°C. What is the MW of Y? 8.0°C/molat, fpe 70.00) . When 0 650g of compound Wis dissolved in 27.8g biphenyl (K fp of the solution is 68.44'C. What is the MW of W? 6. A compound has the following composition: C=93.75% and H=6.25%. When 0.320g of the compound is dissolved in 10 og benzene ( K 5.12"C/molal, fpe 5.50°C) the solution has a lot 4 20'C. Calculate a) the empirical formula and b) the molecular formula

Answer 1

Q1)

Mass of Quinone dissolved = 20.0 g

Molar Mass of Quinone (C₆H₄O₂) = 108.1 g/mol

Moles of Quinone = Mass / Molar Mass= (20.0 g) / (108.1 g/mol) = 0.185 mol

Weight of Solvent (Acentone) = 100 g = 100 g * (0.001 kg/g) = 0.100 kg

Molality, b = Moles / Weight of Solvent in kg = 0.185 mol / 0.100 kg = 1.850 molal

Ebullioscopic constant, K_b = 1.67 ^oC/molal

Thus we know that the boiling point elevation () is calculated using the formula,

T

$\Delta T = b\cdot K_b = (1.850 molal)\times(1.67^oC/molal)=3.09 ^oC$

Thus the new boiling point = Original + Change = 56.5 ^oC + 3.09 ^oC= 59.59 ^oC (Answer)

2)

Here as well, we will use the same process except that the presence of Solute causes a depression in freezing point and here all the variables except the Molecular Weight is given. Thus, we will proceed in reverse order.

$\Delta T = b\cdot K_f \\ \Rightarrow (5.50 ^oC-3.27^o C)= b\times (5.12^oC/molal)\\ \Rightarrow b = \frac{2.23 ^oC}{5.12^oC/molal}=0.4355 molal$

Mass of solvent (benzene) = 25 g = 25 g * (0.001 kg/g) = 0.025 kg

Moles of Solute= Molality * Mass of Solvent = 0.4355 molal * 0.025 kg = 0.01089 mol

Mass of solute dissolved = 1.286 g

Molar Mass of Solute, MW = Mass / Moles = (1.286 g / 0.01089 mol) = 118.1 g/mol (Answer)