QJEEM21Ch6aca: JEE Main 2021 Chemistry - Class 12 JEEM21Ch6aca Solution: 40 g of glucose (Molar mass = 180) is mixed with 200 mL of…

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QJEEM21Ch6aca: JEE Main 2021 Chemistry - Class 12 JEEM21Ch6aca Solution: 40 g of glucose (Molar mass = 180) is mixed with 200 mL of…

40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is __________ K. (Nearest integer) [Given : K_f = 1.86 K kg mol^{$-$ 1}; Density of water = 1.00 g cm^{$-$ 3}; Freezing point of water = 273.15 K]

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Accepted Answer

Molality =

\frac{(\frac{40}{180}) m o l}{0.2 K g} = (\frac{10}{9})

molal

\Rightarrow Δ T_{f} = T_{f} - {T_{f}}^{'} = 1.86 \times \frac{10}{9}

\Rightarrow {T_{f}}^{'} = 273.15 - 1.86 \times \frac{10}{9}

= 271.08 K

≃

271 K (nearest - integer)

QJEEM21Ch6aca: JEE Main 2021 Chemistry - Class 12 JEEM21Ch6aca Solution: 40 g of glucose (Molar mass = 180) is mixed with 200 mL of…

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