1 g of AB₂ is dissolved in 50 g solvent such that ∆Tf = 0.689. When 1 g AB is dissolved in 50 g of same solvent, ∆Tf is 1.176. Find molar mass of AB2. Kf = 5 K kg/mol. AB₂ and AB are non electrolyte. (Report to nearest integer)

The correct answer is 145g.

Step-by-Step Explanation:

We are given the depression in freezing point ($\Delta T_f$) for solutions of two non-electrolytes, AB₂ and AB, in the same solvent. We need to find the molar mass of AB₂.

The relation between depression in freezing point and molality of a non-electrolyte solute is given by the formula:
$\Delta T_f=K_f\times m$
where:
• $\Delta T_f$ is the depression in freezing point,
• $K_f$ is the molal depression constant (cryoscopic constant) of the solvent,
• $m$ is the molality of the solution.

Molality ($m$) is defined as the number of moles of solute dissolved per kilogram of solvent:
$m=\frac{w_2\times 1000}{M_2\times w_1}$
where:
• $w_2$ is the mass of the solute in grams (1 g),
• $M_2$ is the molar mass of the solute in g/mol,
• $w_1$ is the mass of the solvent in grams (50 g).

Substituting the molality formula into the depression in freezing point equation:
$\Delta T_f=\frac{K_f\times w_2\times 1000}{M_2\times w_1}$

Now, let us substitute the given values for the solute AB₂:
• Mass of AB₂ ($w_2$) = 1 g
• Mass of solvent ($w_1$) = 50 g
• Cryoscopic constant ($K_f$) = 5 K kg/mol
• Depression in freezing point ($\Delta T_f$) = 0.689 K

Let $M_{{\mathrm{AB}}_2}$ be the molar mass of AB₂. Substituting these values into the equation:
$0.689=\frac{5\times 1\times 1000}{M_{{\mathrm{AB}}_2}\times 50}$

Simplify the expression on the right-hand side:
$0.689=\frac{5000}{50\times M_{{\mathrm{AB}}_2}}$
$0.689=\frac{100}{M_{{\mathrm{AB}}_2}}$

Rearranging the equation to solve for the molar mass of AB₂ ($M_{{\mathrm{AB}}_2}$):
$M_{{\mathrm{AB}}_2}=\frac{100}{0.689}$
$M_{{\mathrm{AB}}_2}\approx 145.14\text{g/mol}$

Rounding to the nearest integer as requested, we get:
$M_{{\mathrm{AB}}_2}=145\text{g/mol}$