83 g of ethylene glycol is dissolved in 625 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$: $(2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g \ mol^{-1}$.$2$. Calculate

Vedclass · Accepted Answer

$269$

Vedclass · Answer

(B) $1$. Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$: $(2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g \ mol^{-1}$.$2$. Calculate the number of moles of solute: $n = \frac{83 \ g}{62 \ g \ mol^{-1}} \approx 1.3387 \ mol$.$3$. Calculate the molality $(m)$ of the solution: $m = \frac{n 	ext{ (mol)}}{W_{	ext{solvent}} 	ext{ (kg)}} = \frac{1.3387 \ mol}{0.625 \ kg} = 2.1419 \ mol \ kg^{-1}$.$4$. Calculate the depression in freezing point $(\Delta T_f)$: $\Delta T_f = K_f 	imes m = 1.86 	imes 2.1419 \approx 3.984 \ K \approx 4 \ K$.$5$. Calculate the freezing point of the solution $(T_f)$: $T_f = T_f^{\circ} - \Delta T_f = 273 \ K - 4 \ K = 269 \ K$.

Similar Questions

$1.00 \, g$ of a non-electrolyte solute dissolved in $50 \, g$ of benzene lowered the freezing point of benzene by $0.40 \, K$. The freezing point depression constant of benzene is $5.12 \, K \, kg \, mol^{-1}$. Find the molar mass of the solute.

At $1 \ atm$ pressure,which of the following solutions will have the highest freezing point?

What is molal depression constant? Derive an equation.

What is the mass of solute having molar mass $60 \ g \ mol^{-1}$ when dissolved in $98 \ g$ of solvent decreases its freezing point by $0.2 \ K$ (in $g$)? (The numerical value of cryoscopic constant of solvent is $1.71 \ K \ kg \ mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module