The freezing point of a solution containing 36 g of a compound (empirical formula: CH 2 O ) in 1.20 kg of water is found to be 0.93° C. The molecular formula of the solute is (Kf. of water is .1.86 K Kgmol -1)

Question

The freezing point of a solution containing 36 g of a compound (empirical formula: CH 2 O ) in 1.20 kg of water is found to be 0.93° C. The molecular formula of the solute is (Kf. of water is .1.86 K Kgmol -1) (A) C H2 O (B) C 3 H 6 O 3 (C) C 4 H 8 O 4 (D) C 2 H 4 O 2

Tardigrade Admin · Accepted Answer

Δ f=Kf × m 0.93 K=1.86 × (36 / M/1.20) M=(1.86 × 36/0.93 × 1.2)=60 g mol -1 n=( text Molar mass / text Empirical formula mass ) n=(60/30)=2 Hence molecular formula =( CH 2 O ) × 2= C 2 H 4 O 2

Q. The freezing point of a solution containing $36 g$ of a compound (empirical formula: $C H_{2} O$ ) in $1.20 k g$ of water is found to be $0.9 3^{\circ} C$ . The molecular formula of the solute is $(K_{f}$ of water is $1.86 KK g m o l^{- 1})$

$C H_{2} O$

$C_{3} H_{6} O_{3}$

$C_{4} H_{8} O_{4}$

$C_{2} H_{4} O_{2}$

$Δ f = K_{f} \times m$
$0.93 K = 1.86 \times \frac{36/ M}{1.20}$
$M = \frac{1.86 \times 36}{0.93 \times 1.2} = 60 g m o l^{- 1}$
$n = \frac{Molar mass}{Empirical formula mass}$
$n = \frac{60}{30} = 2$
Hence molecular formula $= (C H_{2} O) \times 2 = C_{2} H_{4} O_{2}$

Q. The freezing point of a solution containing 36g of a compound (empirical formula: CH2​O ) in 1.20kg of water is found to be 0.93∘C. The molecular formula of the solute is (Kf​ of water is 1.86KKgmol−1)

CH2​O

C3​H6​O3​

C4​H8​O4​

C2​H4​O2​