A solution containing 1.8 g of a compound (empirical formula CH2O ) in 40 g of water is observed to freeze at -0.465° C. The molecular formula of the compound is ( kf of water = 1.86 text kg K mol-1 )

Question

A solution containing 1.8 g of a compound (empirical formula CH2O ) in 40 g of water is observed to freeze at -0.465° C. The molecular formula of the compound is ( kf of water = 1.86 text kg K mol-1 ) (A)  C2H4O2  (B)  C3H6O3  (C)  C4H8O4  (D)  C5H10O5

Tardigrade Admin · Accepted Answer

Δ Tf=Kf× m ∵ Δ Tf=Tfo-Tf =0° -(-0.465° )=0.465° C ∴ 0.465=1.86× (1.8/M)× (1000/40) or M=180 empirical formula weight =12+1× 2+16 =30 ∵ n=(molecular text weight/empirical text formula text weight)=(180/30)=6 ∴ Molecular formula =(CH2O)6 =C2H12O6 Thus, molecular formula of compound =C6H12O6

Q. A solution containing $1.8 g$ of a compound (empirical formula $C H_{2} O$ ) in $40 g$ of water is observed to freeze at - $0.46 5^{\circ} C$ . The molecular formula of the compound is ( $k_{f}$ of water = $1.86 k g K m o l^{- 1}$ )

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

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

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

$C_{5} H_{10} O_{5}$

$C_{6} H_{12} O_{6}$

Q. A solution containing 1.8g of a compound (empirical formula CH2​O ) in 40g of water is observed to freeze at -0.465∘C. The molecular formula of the compound is ( kf​ of water = 1.86kgKmol−1 )

C2​H4​O2​

C3​H6​O3​

C4​H8​O4​

C5​H10​O5​