TS EAMCET 4-Aug-2021 Shift 1 Question Paper

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.