When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be -0.93° C ( K f( H 2 O )=1.86 K kg . mol -1 ). The number (n) of benzoic acid molecules associated (assuming 100 % association) is

Question

When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be -0.93° C ( K f( H 2 O )=1.86 K kg . mol -1 ). The number (n) of benzoic acid molecules associated (assuming 100 % association) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ T f = i × k f × m 0-(-0.93)= i × 1.86 × (12.2/122 × 100) × 1000 i =(0.93/1.86)=0.5 i =1+((1/ n )-1) α (1/2)=1+((1/ n )-1) × 1 n =2

Q. When $12.2 g$ of benzoic acid is dissolved in $100 g$ of water, the freezing point of solution was found to be $- 0.9 3^{\circ} C (K_{f} (H_{2} O) = 1.86 K k g$ $m o l^{- 1}$ ). The number (n) of benzoic acid molecules associated (assuming $100%$ association) is___

$Δ T_{f} = i \times k_{f} \times m$

$0 - (- 0.93) = i \times 1.86 \times \frac{12.2}{122 \times 100} \times 1000$

$i = \frac{0.93}{1.86} = 0.5$

$i = 1 + (\frac{1}{n} - 1) α$

$\frac{1}{2} = 1 + (\frac{1}{n} - 1) \times 1$

$n = 2$

Q. When 12.2g of benzoic acid is dissolved in 100g of water, the freezing point of solution was found to be −0.93∘C(Kf​(H2​O)=1.86Kkg mol−1 ). The number (n) of benzoic acid molecules associated (assuming 100% association) is___

ΔTf​=i×kf​×m 0−(−0.93)=i×1.86×122×10012.2​×1000 i=1.860.93​=0.5 i=1+(n1​−1)α 21​=1+(n1​−1)×1 n=2

Q. When $12.2 g$ of benzoic acid is dissolved in $100 g$ of water, the freezing point of solution was found to be $- 0.9 3^{\circ} C (K_{f} (H_{2} O) = 1.86 K k g$ $m o l^{- 1}$ ). The number (n) of benzoic acid molecules associated (assuming $100%$ association) is___

$Δ T_{f} = i \times k_{f} \times m$

$0 - (- 0.93) = i \times 1.86 \times \frac{12.2}{122 \times 100} \times 1000$

$i = \frac{0.93}{1.86} = 0.5$

$i = 1 + (\frac{1}{n} - 1) α$

$\frac{1}{2} = 1 + (\frac{1}{n} - 1) \times 1$

$n = 2$