150 g of acetic acid was contaminated with 10.2 g ascorbic acid ( C 6 H 8 O 6) to lower down its freezing point by ( x × 10-1)° C. The value of x is (Nearest integer) [Given K f =3.9 K kg mol -1; Molar mass of ascorbic acid =176 g mol -1 ]

Question

150 g of acetic acid was contaminated with 10.2 g ascorbic acid ( C 6 H 8 O 6) to lower down its freezing point by ( x × 10-1)° C. The value of x is (Nearest integer) [Given K f =3.9 K kg mol -1; Molar mass of ascorbic acid =176 g mol -1 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

150 g CH 3 COOH 10.2 g text ascorbic acid ⇒ 0.058 text moles Δ T f =( x × 10-1) ° C Δ T f = k f ⋅ text molality =3.9 × (0.058/150) × 1000 =1.5° C =15 × 10-1 ° C

Q. $150 g$ of acetic acid was contaminated with $10.2 g$ ascorbic acid $(C_{6} H_{8} O_{6})$ to lower down its freezing point by $(x \times 1 0^{- 1})^{\circ} C$ . The value of $x$ is (Nearest integer) [Given $K_{f} = 3.9 K k g m o l^{- 1}$ ; Molar mass of ascorbic acid $= 176 g m o l^{- 1}$ ]

$150 g C H_{3} COO H$
$10.2 g ascorbic acid \Rightarrow 0.058 moles$
$Δ T_{f} = (x \times 1 0^{- 1})^{\circ} C$
$Δ T_{f} = k_{f} \cdot molality$
$= 3.9 \times \frac{0.058}{150} \times 1000$
$= 1. 5^{\circ} C$
$= 15 \times 1 0^{- 1}^{\circ} C$

Q. 150g of acetic acid was contaminated with 10.2g ascorbic acid (C6​H8​O6​) to lower down its freezing point by (x×10−1)∘C. The value of x is (Nearest integer) [Given Kf​=3.9Kkgmol−1; Molar mass of ascorbic acid =176gmol−1 ]

150gCH3​COOH 10.2g ascorbic acid ⇒0.058 moles ΔTf​=(x×10−1)∘C ΔTf​=kf​⋅ molality =3.9×1500.058​×1000 =1.5∘C =15×10−1∘C

Q. $150 g$ of acetic acid was contaminated with $10.2 g$ ascorbic acid $(C_{6} H_{8} O_{6})$ to lower down its freezing point by $(x \times 1 0^{- 1})^{\circ} C$ . The value of $x$ is (Nearest integer) [Given $K_{f} = 3.9 K k g m o l^{- 1}$ ; Molar mass of ascorbic acid $= 176 g m o l^{- 1}$ ]

$150 g C H_{3} COO H$
$10.2 g ascorbic acid \Rightarrow 0.058 moles$
$Δ T_{f} = (x \times 1 0^{- 1})^{\circ} C$
$Δ T_{f} = k_{f} \cdot molality$
$= 3.9 \times \frac{0.058}{150} \times 1000$
$= 1. 5^{\circ} C$
$= 15 \times 1 0^{- 1}^{\circ} C$