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3 years ago

Cl2 + 2OH− → Cl− + ClO− + H2O

1 answer:

4 0

Cl2(g) -------> Cl-(aq) + ClO-(aq)

2e- + Cl2(g) -------> 2Cl-(aq) [reduction]

4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
______________________________________...
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)

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Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde

pochemuha

Solution :

We know that :

$\Delta T_f = k_f.m$ and $m=\frac{w_2}{m_2 \times w_1}$

Then, $\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$ ..................(1)

Where,

$w_1$ = amount of solvent (in kg)

$w_2$ = amount of solute (in kg)

$m_2$ = molar mass of solute (g/mole)

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Given :

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Then putting this values in the equation is (1),

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