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  4. 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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Q. $40\, g$ of glucose (Molar mass $=180$ ) is mixed with $200\, mL$ of water. The freezing point of solution is _______$K$. (Nearest integer)
$\left[\right.$ 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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Solution:

molality $=\frac{\left(\frac{40}{180}\right) mol }{0.2\, Kg }=\left(\frac{10}{9}\right) molal$
$\Rightarrow \Delta T _{ f }= T _{ f }- T _{ f }^{\prime}=1.86 \times \frac{10}{9}$
$\Rightarrow T _{ f }^{\prime}=273.15-1.86 \times \frac{10}{9}$
$=271.08\, K$
$\simeq 271\, K$ (nearest-integer)