CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Calculate the temperature at which the solution of $$ 54 \mathrm{g} $$ glucose present in $$ 250 \mathrm{g} $$ water will be freezed. $$ \left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right) $$


Solution

Given that:
Weight of solute $$ \left(\mathrm{W}_{\mathrm{B}}\right)=54 \mathrm{g} $$

Molecular weight of solute $$ \left(\mathrm{M}_{\mathrm{B}}\right)=180 \mathrm{g} $$ molWeight of solvent $$ \left(\mathrm{W}_{\mathrm{A}}\right)=250 \mathrm{g} $$

$$ \mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} $$

Depression in freezing point $$ \Delta T_{f} $$

$$ =\dfrac{K_{f} \times W_{B} \times 1000}{M_{B} \times W_{A}} $$

$$ =\dfrac{1.86 \times 54 \times 1000}{180 \times 250}=2.23 \mathrm{K} $$

Depression in freezing point $$ =2.23 \mathrm{K} $$

Freezing point of solution $$ =? $$

Freezing point of water $$ =273.15 \mathrm{K} $$

$$ \left(\Delta \mathrm{T}_{\mathrm{f}}\right)= $$ Freezing point of water - Freezing point of solution

Freezing point of solution $$ =273.15-2.23=270.92 K $$

Chemistry
NCERT
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More



footer-image