As we know, the relation between molarity (M) and molality (m). m=
1000×M
(1000×d)−(M×m
. . . (i) here, m= molality m=
(1000×d)
m= molality M= molarity ⇒M=2 (given) d= density of solution in g∕mL(d=1.11g∕mL) Molar mass of ethylene glycol ⇒62.07g∕mol Put all values in equation number (i), we get m=
1000×2
(1000×1.11)−(2×62.07)
g∕mol =
2000
1110−124.14
=
2000
985.86
=2.028≡−2.05 Hence, molality of the solution is 2.05m.