An antifreeze solution is prepared from 222.6 | vedclass

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

Question

(A) The molar mass of ethylene glycol $[C_2H_4(OH)_2]$ is $(2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g/mol$. Molality $(m)$ is defined as th

Vedclass · Accepted Answer

$17.95$

Vedclass · Answer

(A) The molar mass of ethylene glycol $[C_2H_4(OH)_2]$ is $(2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g/mol$. Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Moles of ethylene glycol $= \frac{222.6 \ g}{62 \ g/mol} = 3.59 \ mol$. Mass of solvent (water) $= 200 \ g = 0.2 \ kg$. Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{3.59 \ mol}{0.2 \ kg} = 17.95 \ m$.

An antifreeze solution is prepared from $222.6 \ g$ of ethylene glycol $[C_2H_4(OH)_2]$ and $200 \ g$ of water. What is the molality of this solution (in $m$)?

Similar Questions

If $98 \, mg$ of $H_2SO_4$ is present in $100 \, g$ of solution,what is the molarity of the solution? (Given: density of solution = $1.25 \, g/mL$)

When is the parts per million $(ppm)$ unit used?

How many grams of $NaOH$ must be dissolved in $500 \, g$ of solution to obtain a $10 \% \, w/w$ concentration?

$0.1 \ mol$ of $NaCl$ is dissolved in $100 \ g$ of water. The mole fraction of $NaCl$ is

Calculate the molarity $(M)$ of a $63 \% \ w/w$ $HNO_3$ solution if the density is $5.4 \ g/mL$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module