Question

Density of 1 M solution of glucy is 1.18g for water is 1.86km. find freezing point of solution.​

Answer 1

Question:

Density of 1 M solution of glucose is 1.18 $g/cm^3$. $K_f$ of  H₂O is 1.86 $Km^-1$ Find freezing point of the solution.

Answer:

$\bigstar$ Given:

⇒ 1 Molar solution

⇒ Density of 1 M solution = 1.18 $g/cm^3$

⇒ $K_f$ of H₂O = 1.86 $Km^-1$

$\bigstar$ To Find:

⇒ Freezing point of the Solution $T_f$

$\bigstar$ Solution:

Depression in Freezing point is directly proportional to molality (m)  i.e.,

$\green{\boxed{{\boxed{\Delta\;T_f\;\; \alpha \;\; m}}}}$

$\Delta\;T_f = K_f * m$

$\Delta T_f = T^0_f \;- \;T_f$

Here,

$T^0_f$ represents the freezing point of pure solvent.

$T_f$ represent the freezing point when non-volatile solute is dissolved in it.

Therefore,

$T^0_f \;- \;T_f = K_f * m$

$0^0C - T_f = 1.86Km^-1 * m\;\;==>\; 1$

Since, Freezing point of pure solvent is at $0^0C$

$K_f$ of H₂O = 1.86 $Km^-1$

Density is given by the formula,

${\boxed{{density \;of \; soln =\dfrac {mass \;of\; soln}{vol\; of \;soln} }}}$

$1.18 \;gm/cc = \dfrac{mass\;of\;solution}{1000\;cc}$

∴ Mass of solution = 1180 gm

Weight of Solute = Molecular Mass of Glucose (C₆H₁₂O₆)

Weight of Solute = 6*12 + 12*1 + 6*16

Weight of Solute = 96+12+72

∴ Weight of Solute = 180 gm

Therefore, the weight of solvent = 1180 - 180 = 1000 g = 1 Kg

Weight of solvent = 1 Kg

To Find Molality (m) we use to formula,

$\blue{\boxed{m = \dfrac{number \;of\;moles\;of\;solute}{weight\; of\;solvent\;(in \;Kg)}}}$

$m = \dfrac{1\;mole}{1\;Kg}$

$\pink{\boxed{m = 1 }}$

Substituting this in equation 1, We get

$-T_f = 1.86 * 1$

$-T_f = 1.86$

Therefore,

$\huge{\green{\boxed{\orange{{T_f = -1.86^0C}}}}}$

The Freezing point is -1.86°C