(C) The freezing point of a solution is lowered by the addition of a solute (depression in freezing point). $CaCl_2$ is preferred because the eutectic

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Eutectic mixture of $CaCl_2/H_2O$ freezes at $-55 \ ^oC$ while that of $NaCl/H_2O$ freezes at $-18 \ ^oC$

(C) The freezing point of a solution is lowered by the addition of a solute (depression in freezing point). $CaCl_2$ is preferred because the eutectic mixture of $CaCl_2$ and $H_2O$ has a much lower freezing point $(-55 \ ^oC)$ compared to the eutectic mixture of $NaCl$ and $H_2O$ $(-18 \ ^oC)$. This allows $CaCl_2$ to effectively melt ice at significantly lower temperatures.

Class 12 Chemistry Solutions Depression of freezing point of the solvent

Difficult

$CaCl_2$ is preferred over $NaCl$ for clearing ice on roads,particularly in very cold countries. This is because:

A
$CaCl_2$ is less soluble in $H_2O$ than $NaCl$
B
$CaCl_2$ is hygroscopic but $NaCl$ is not
C
Eutectic mixture of $CaCl_2/H_2O$ freezes at $-55 \ ^oC$ while that of $NaCl/H_2O$ freezes at $-18 \ ^oC$
D
$NaCl$ makes the road slippery but $CaCl_2$ does not

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Depression of freezing point of the solvent

The freezing point of $1 \%$ solution of lead nitrate in water will be

Medium

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Ethylene glycol is used as an antifreeze in cold climates. The mass of ethylene glycol $(C_2H_6O_2)$ that should be added to $4 \ kg$ of water to prevent it from freezing at $-6 \ ^oC$ is ......... $g$.
($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$,and molar mass of ethylene glycol $= 62 \ g \ mol^{-1}$)

DifficultAIEEE 2011

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$1.8 \ g$ of glucose (molar mass $180 \ g \ mol^{-1}$) is dissolved in $0.1 \ kg$ of water. The freezing point of the solution (in $^{\circ}C$) is ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)

MediumAP EAMCET 2022

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The freezing point of a $5\%$ (by mass) aqueous solution of sucrose is $271 \, K$ and the freezing point of pure water is $273.15 \, K$. The freezing point of a $5\%$ (by mass) aqueous solution of glucose is .......... $K$.

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How much glucose $(molecular \ weight = 180 \ g/mol)$ should be added to $200 \ g \ H_2O$ so that when the solution is cooled to $-0.5^{\circ}C$,$14 \ g$ of ice separates out of the solution: [$K_f = 1.86 \ K \ kg/mol$ and melting point of $H_2O = 0^{\circ}C$] (in $g$)

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$CaCl_2$ is preferred over $NaCl$ for clearing ice on roads,particularly in very cold countries. This is because:

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The freezing point of $1 \%$ solution of lead nitrate in water will be

$1.8 \ g$ of glucose (molar mass $180 \ g \ mol^{-1}$) is dissolved in $0.1 \ kg$ of water. The freezing point of the solution (in $^{\circ}C$) is ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)

The freezing point of a $5\%$ (by mass) aqueous solution of sucrose is $271 \, K$ and the freezing point of pure water is $273.15 \, K$. The freezing point of a $5\%$ (by mass) aqueous solution of glucose is .......... $K$.

How much glucose $(molecular \ weight = 180 \ g/mol)$ should be added to $200 \ g \ H_2O$ so that when the solution is cooled to $-0.5^{\circ}C$,$14 \ g$ of ice separates out of the solution: [$K_f = 1.86 \ K \ kg/mol$ and melting point of $H_2O = 0^{\circ}C$] (in $g$)

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