A certain mass of substance- when dissolved in 100 g C 6 H 6- lowers the freezing point by 1-28- C- The same mass of solute dissolved in 100 g water lowers the freezing point by 1-40- C- If the substance has normal molecular weight in benzene and is completely ionized in water- into how many ions does it dissociate in water- K f for H 2 O and C 6 H 6 are 1-86 and 5-12 K kg mol -1-

Δ T_f = i× m× K_f For benzene and water respectively: 1.28 = 1 × 5.12 ×n/0.1 1.40 = i × 1.86 ×n/0.1 So, i = 3. ...

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Standard X

Chemistry

Solubility

A certain mas...

Question

A certain mass of substance, when dissolved in 100 g $C_{6} H_{6}$ , lowers the freezing point by ${1.28}^{\circ} C$ . The same mass of solute dissolved in 100 g water lowers the freezing point by ${1.40}^{\circ} C$ . If the substance has normal molecular weight in benzene and is completely ionized in water, into how many ions does it dissociate in water? $K_{f}$ for $H_{2} O$ and $C_{6} H_{6}$ are 1.86 and 5.12 K kg $m o l^{- 1}$ .

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Solution

$Δ T_{f} = i \times m \times K_{f}$
For benzene and water respectively:
$1.28 = 1 \times 5.12 \times \frac{n}{0.1} 1.40 = i \times 1.86 \times \frac{n}{0.1}$

So, $i = 3$ .

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ΔTf=i×m×Kf For benzene and water respectively: 1.28=1×5.12×n0.11.40=i×1.86×n0.1 So, i=3.

$Δ T_{f} = i \times m \times K_{f}$
For benzene and water respectively:
$1.28 = 1 \times 5.12 \times \frac{n}{0.1} 1.40 = i \times 1.86 \times \frac{n}{0.1}$

So, $i = 3$ .