$40 g$ of glucose (Molar mass $= 180)$ is mixed with $200 m L$ of water. The freezing point of solution is ________ $K$ . (Nearest integer)
$[Given : K_{f} = 1.86 K m o l^{- 1} Density of water = 1.00 g c m^{- 3}; Freezing point water = 273.15 K]$

271.0

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271

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271.00

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Solution

Freezing point depression of a solution is given by,
$△ T_{f} = i \times K_{f} \times m$
where,
$△ T_{f}$ is the depression in freezing point.
$K_{f}$ is molal depression constant.
m is molality of solution.
$Van't hoff factor, i = 1 for glucose)$
$Molality (m) = \frac{\frac{w_{solute}}{M} \times 1000}{W_{solvent}}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute

$Δ T_{f} = 1 \times 1.86 \times \frac{40 \times 1000}{180 \times 200} (\frac{Mass of water = 200 g}{a s d = 1 g / m L})$

$= 2.06$

$∴$ Freezing point $= T_{o} - Δ T_{f}$
$T_{o}$ is freezing point of water
$= 273.15 - 2.06$

$= 271.09 K$

$= 271 K$

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Q. A

10

% solution (by mass) of sucrose in water has a freezing point of

269.15 K

. Calculate the freezing point of

10

% glucose in water if the freezing point of pure water is

273.15 K

.
(Given : molar mass of sucrose

= 342 g m o l^{-}

, Molar mass of glucose

= 180 g m o l^{- 1}

)

(a) A $10 %$ solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of $10 %$ glucose in water if the freezing point of pure water is 273.15 K.
given:
(Molar mass of sucrose = 342 g $m o l^{- 1}$ )
(Molar mass of glucose = 180 g $m o l^{- 1}$ )
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass.