Question

The freezing point of the solution containing 0.3 g of acetic acid and 30 g of benzene is lowered by 0.45 degree c then van't hoff factor is (Kf of benene is 5.12Kkg/mol)

Answer 1

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Given:

Answer 2

The freezing point of the solution containing 0.3 g of acetic acid and 30 g of benzene is lowered by $0.45^0C$ then van't hoff factor is 0.68

Explanation:

Depression in freezing point = $0.45^oC$

Mass of acetic acid (solute) = 0.3 g

Mass of benzene (solvent) = 30 g = 0.03 kg    (1kg = 1000 g)

Formula used :  

$\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of acetic acid}}{\text{Molar mass of benzene}\times \text{Mass of benzene in kg}}$

where,

i = Van't Hoff factor = ?

$K_f$ = freezing point constant = 5.12 Kkg/mole

m = molality

Now put all the given values in this formula, we get

$0.45=i\times (5.12K.kg/mole)\times \frac{0.3g}{78g/mol\times 0.03kg}$

$i=0.68$

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