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Solutions — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistrySolutions4 Marks
Question
Read the passage given below and answer the following questions
Few colligative properties are:
  1. Relative lowering of vapour pressure: depends only on molar concentration of solute (mole fraction) and independent of its nature.
  2. Depression in freezing point: it is proportional to the molal concentration of solution.
  3. Elevation of boiling point: it is proportional to the molal concentration of solute.
  4. Osmotic pressure: it is proportional to the molar concentration of solute
A solution of glucose is prepared with $0.052$ g at glucose in $80.2$ g of water. $\left(\mathrm{KJ}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right.$ and $\left.\mathrm{K}_{\mathrm{b}}=5.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$ The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Molality of the given solution is.
  1. $0.0052m$
  2. $0.0036m$
  3. $0.0006m$
  4. $1.29m$
  1. Boiling point for the solution will be.
  1. $373.05K$
  2. $373.15K$
  3. $373.02K$
  4. $373.02K$
  1. The depression in freezing point of solution will be.
  1. $0.0187K$
  2. $0.035K$
  3. $0.082K$
  4. $0.067K$
  1. Mole fraction of glucose in the given solution is.
  1. $6.28 \times 10^{-5}$
  2. $6.28 \times 10^{-4}$
  3. $0.00625$
  4. $0.00028$
  1. If same amount of sucrose $(C_{12} H_{22} O_{11})$ is taken instead of glucose, then.
  1. Elevation in boiling point will be higher.
  2. Depression in freezing point will be higher.
  3. Depression in freezing point will be lower.
  4. Both $(a)$ and $(b).$
✓

Answer

  1. (b) $0.0036m$
Explanation:

$\text{m}=\frac{0.052}{180}\times\frac{1000}{80.2}=0.0036$
  1. (c) $373.02K$
Explanation:

$\Delta\text{T}_\text{b}=\text{k}_\text{b}\times\text{m}=5.2\times0.0036=0.0187\ \text{K}$

$T_b = 373 + 0.0187 = 373.0187 K  \approx  373.02 K$
  1. (d) $0.067 K$
Explanation:

$\Delta\text{T}_\text{f}=\text{k}_\text{f}\times\text{m}=1.86\times0.0036=0.067\ \text{K}$
  1. (a) $6.28 \times 10^{-5}$​​​​​​​
Explanation:

Moles of water $\frac{80.2}{18}=4.455$

Mole fraction of glucose $=\frac{0.00028}{4.45+0.00028}=6.28\times10^{-5}$'
  1. (c) Depression in freezing point will be lower.
Explanation:

Depression in freezing point or elevation in boiling point is proportional to molarity, which is proportional to number of moles. For same amount, higher the molar mass of solute, lower will be number of moles. Hence, lower will be the colligative property.

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