Question

calculate the freezing point of a solution containing 60g of glucose (molar mass = 180g mol-1)in 250 g of water.(kf of water = 1•86 k kg mol-1)

Answer 1

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Answer 2

Given:

Weight of water ($w_1$) = 250g

Weight of glucose ($w_2$) = 60g

Molar mass of water ($M_2$) = 180g/mol

$k_f$ of water = 1.86 k Kg/mol

To find:

Freezing point of solution = ?

Formula to be used:

Δ$T_f$ = $k_fm$

Δ$T_f$ = $k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

Calculation:

Δ$T_f$ = $k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

Δ$T_f$ = $1.86\times \frac{60\times 1000}{180\times 250}$

Δ$T_f$ = $\frac{1.86\times 600}{18\times 25}$

Δ$T_f$ = $\frac{1116}{450}$

Δ$T_f$ = 2.48K

∵ Δ$T_f$ = $T_{solvent} - T_{Solution}$

$T_{solution} = T_{solvent} -$ Δ$T_f$

$T_{solution} =$ 273.15 k - 2.48 k = 270.67 k

Conclusion:

The freezing point of the given solution is calculated as 270.667K

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