Question

Calculate the amount of KCL which must be added to 1kg of water so the freezing point is depressed by 2k

Answer 1

$change\ in\ freezing\ point,\ \Delta T_f=k_f \times m\\ \\ \Rightarrow m= \frac{\Delta T_f}{k_f} \\ \\ \Rightarrow m= \frac{2}{1.86} =1.075\ mole/kg\\ \\m= \frac{no.\ of\ moles\ of\ solute}{mass\ of\ solvent}= \frac{n}{mass\ of\ water} \\ \\ \Rightarrow 1.075= \frac{n}{1} \\ \\ \Rightarrow n=1.075\ moles\\ \\ \therefore amount\ of\ KCl\ needed\ is\ 1.075\ mole.$

Answer 2

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$change\ in\ freezing\ point,\ \Delta T_f=k_f \times m\\ \\ \Rightarrow m= \frac{\Delta T_f}{k_f} \\ \\ \Rightarrow m= \frac{2}{1.86} =1.075\ mole/kg\\ \\m= \frac{no.\ of\ moles\ of\ solute}{mass\ of\ solvent}= \frac{n}{mass\ of\ water} \\ \\ \Rightarrow 1.075= \frac{n}{1} \\ \\ \Rightarrow n=1.075\ moles\\ \\ \therefore amount\ of\ KCl\ needed\ is\ 1.075\ mole.$

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