Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2° C. The percentage association of solute A in water, is [Given: Molar mass of A =93 g mol -1. Molal depression constant of water is 1.86 K kg mol -1 ]

Question

Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2° C. The percentage association of solute A in water, is [Given: Molar mass of A =93 g mol -1. Molal depression constant of water is 1.86 K kg mol -1 ] (A) 50 % (B) 60 % (C) 70 % (D) 80 %

Tardigrade Admin · Accepted Answer

Δ T = i k f × m 0.2= i × 1.86 × (0.7/93) × (1000/42) i =(0.2 × 93 × 6/1.86 × 100) i =0.60 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/0d94081836814b9a17c8156f8dca9dc9-.png" /> i =1-α+(α/2) i =1-(α/2) 1-(α/2)=0.60 1-0.60=(α/2) α=0.80

$50%$

$60%$

$70%$

$80%$

$Δ T = i k_{f} \times m$
$0.2 = i \times 1.86 \times \frac{0.7}{93} \times \frac{1000}{42}$
$i = \frac{0.2 \times 93 \times 6}{1.86 \times 100}$
$i = 0.60$

$i = 1 - α + \frac{α}{2}$
$i = 1 - \frac{α}{2}$
$1 - \frac{α}{2} = 0.60$
$1 - 0.60 = \frac{α}{2}$
$α = 0.80$

Q. Solute A associates in water. When 0.7g of solute A is dissolved in 42.0g of water, it depresses the freezing point by 0.2∘C. The percentage association of solute A in water, is [Given : Molar mass of A=93gmol−1. Molal depression constant of water is 1.86Kkgmol−1 ]

50%

60%

70%

80%

ΔT=ikf​×m 0.2=i×1.86×930.7​×421000​ i=1.86×1000.2×93×6​ i=0.60 i=1−α+2α​ i=1−2α​ 1−2α​=0.60 1−0.60=2α​ α=0.80

$50%$

$60%$

$70%$

$80%$

$Δ T = i k_{f} \times m$
$0.2 = i \times 1.86 \times \frac{0.7}{93} \times \frac{1000}{42}$
$i = \frac{0.2 \times 93 \times 6}{1.86 \times 100}$
$i = 0.60$

$i = 1 - α + \frac{α}{2}$
$i = 1 - \frac{α}{2}$
$1 - \frac{α}{2} = 0.60$
$1 - 0.60 = \frac{α}{2}$
$α = 0.80$