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Solutions — Chemistry STD 12 Science — Question

Bihar 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.(KJ = 1.86K kg mol-1 and Kb = 5.2K kg mol-1)
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 × 10-5
  2. 6.28 × 10-4
  3. 0.00625
  4. 0.00028
  1. If same amount of sucrose (C12 H22 O11) 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:

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

Tb = 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 × 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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Aldehydes and ketones having acetyl group $\left(\begin{array}{c}\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \text{CH}_3-\text{C}-\end{array}\right)$ are oxidised by sodium hypohalate (NaOX) or halogen and alkali (X2 + OH-) to corresponding sodium salt having one carbon atoms less than the carbonyl compound and give a haloform.

$\ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\text{CH}_3\xrightarrow[\text{orX}_2+\text{NaOH}]{\text{NaOX}}\\ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\stackrel{-}{\hbox{ O}}\stackrel{+}{\hbox{Na}}+\text{CHX}_3(\text{X = Cl, Br, I})$

Sodium hypoiodite (NaOl) when treated with compounds containing CH3CO-group gives yellow precipitate of iodoform. Haloform reaction does not affect a carbon-carbon double bond present in the compound.

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  1. Which of the following compounds will give positive iodoform test?
  1. Isopropyl alcohol.
  2. Propionaldehyde.
  3. Ethylphenyl ketone.
  4. Benzyl alcohol.
  1. Which of the following compounds is not formed in iodoform reaction of acetone?
  1. CH3COCH2l
  2. lCH2COCH2l
  3. CH3COCHl2
  4. CH3COCl3
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starting compound A corresponds to:

  1. In the following reaction sequence, the correct structures of E, F and G are:

(* implies 13C labelled carbon)

  1. An organic compound 'A' has the molecular formula C3H6O. It undergoes iodoform test. When saturated with HCl it gives 'B' of molecular formula C9H14O. 'A' and 'B' respectively are:
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