Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is

\(\frac{1}{2}\)
\(\frac{2}{3}\)
\(\frac{1}{2} \times \frac{273}{298}\)
\(\frac{1}{3}\)

Solution

1 mol each of the following compounds is dissolved in 1L of solution. Which will have the largest ΔT_b value?

Solution

The value of ΔT_b depends upon two factors 'i' and'm'. It is given that 1 mol of each compound is dissolved in 1 L of solution. Hence molarity is same for all the compounds. Now the van't Hoff factor depends on number of particle i.e. on degree of ionisation which further depends on the bond dissociation energy which is in the order

HI < HBr

i.e., bond dissociation energy of HI is least. Lower the bond dissociation energy, higher is the degree of ionisation and hence higher the number of particles,thus i will be maximum for HI and hence ΔT_b value will be larger for HI.

On mixing 3 g of non – volatile solute in 200 mL of water, its boiling point (100°) becomes 100.52°C. If K_b for water is 0.6K/m then molecular wt. of solute is :

10.5 g mol^-1
12.6 g mol^-1
15.7 g mol^-1
17.3 g mol^-1

Solution

Which has the minimum freezing point ?

One molal NaCl aq. solution
One molal CaCl2 aq. solution
One molal KCl aq. solution
One molal urea aq. solution

Solution

Among the given options, CaCl₂ solution will produce maximum (three) ions per molecule,so it will show minimum freezing point.

The amount of urea to be dissolved in 500 c.c. of water(K = 1.86°C mol^-1) to produce a depression of 0.186°C in the freezing point is :

9 g
6 g
3 g
0.3 g

Solution

A 0.001 molal solution of[Pt(NH₃)₄Cl₄] in water had a freezing point depression of 0.0054°C. If K_f for water is 1.80, the correct formulation for the above molecule is

[Pt(NH₃)₄Cl₃]Cl
[Pt(NH₃)₄Cl₂]Cl₂
[Pt(NH₃)₄Cl]Cl₃
[Pt(NH₃)₄Cl₄]

Solution

ΔT_f =i.Kf.m; 0.0054 = i× 1.8× 0.001
i= 3 so it is[Pt(NH₃)₄Cl₂]Cl₂.

Solution

The correct relationship between the boiling points of very dilute solutions of AlCl₃ (t₁) and CaCl₂ (t₂), having the same molar concentration is

t₁ = t₂
t₁ > t₂
t₂ > t₁
t₂ ≥ t₁

Solution

AlCl₃ furnishes more ions than CaCl₂ and thus possess higher boiling point i.e., T₁ > T₂.

If ? is the degree of dissociation of Na₂SO₄ , the Vant Hoff’s factor (i) used for calculating the molecular mass is

1 + ?
1– ?
1 + 2?
1 – 2?

Solution

In countries nearer to polar region, the roads are sprinkled with CaCl₂. This is

to minimise the snow fall
to minimise pollution
to minimise the accumulation of dust on the road
to minimise the wear and tear of the roads

Solution

CaCl₂ acts as a non-volatile solute and results in depression in freezing point. Thus, snow fall is reduced and prevents blocking of roads in the polar region.

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