A solution of sucrose (molar mass = 342 g mol^-1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water.The freezing point of the solution obtained will be(K_f for water = 1.86 K kg mol^-1).

–0.372°C
–0.520°C
+0.372°C
–0.570°C

Solution

An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?

Addition of NaCl
Addition of Na₂SO₄
Addition of 1.00 molal KI
Addition of water

Solution

When the aqueous solution of one molal KI is diluted with water, concentration decreases, therefore the vapour pressure of the resulting solution increases.

A 0.0020 m aqueous solution of an ionic compound Co(NH₃)₅(NO₂)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (K_f= – 1.86°C/m)

Solution

The molal elevation constant for water is 0.52.What will be the boiling point of 2 molar sucrose solution at 1 atm.pressure? (Assume B.P. of pure water as 100°C)

101.04°C
100.26°C
100.52°C
99.74°C

Solution

If \(\frac{N}{10}\) 50 ml H2SO4, \(\frac{N}{3}\) 30 ml HNO3, \(\frac{N}{2}\) 10 ml HCl is mixed and solution is made to 1L. Then normality of resultant solution is

\(\frac{N}{20}\)
\(\frac{N}{40}\)
\(\frac{N}{50}\)
N

Solution

A solution of urea (mol. mass 56 g mol^-1) boils at 100.18Cat the atmospheric pressure. If K_f and K_b for water are 1.86 and 0.512 K kg mol^-1 respectively, the above solution will freeze at

0.654°C
-0.654°C
6.54°C
-6.54°C

Solution

The freezing point of 1% solution of lead nitrate in water will be :

2°C
1°C
0°C
below 0°C

Solution

Addition of solute to water decreases the freezing point of water (pure solvent).
∴ When 1% lead nitrate (solute)is added to water, the freezing point of water will be below 0°C.

Osmotic pressure of 0.4% urea solution is 1.64 atm and that of 3.42% cane sugar is 2.46 atm. When the above two solutions are mixed, the osmotic pressure of the resulting solution is :

0.82atm
2.46 atm
1.64atm
4.10atm

Solution

Osmotic pressure is a colligative property.Hence resulting osmotic pressure of the solution is given by
?_T= ?₁+ ?₂+ ?₃.........
?_T= 1.64 + 2.46 = 4.10 atm.

What is the freezing point of a solution containing 8.1 g HBr in 100 g water assuming the acid to be 90% ionised ?(K_f for water = 1.86 K kg mol^-1) :

0.85K
– 3.53K
0K
– 0.35K

Solution

Given mass of solute = 8.1 g
Mass of solvent = 100 g
For HBr

Which of the following solutions will exhibit highest boiling point?

0.01 M Na₂SO₄(aq)
0.01 M KNO₃(aq)
0.015M urea (aq)
0.015 M glucose (aq)

Solution

For highest boiling point of a solution, the ΔT_b should be highest,Since, ΔT_b is directly proportional to the molality of solution ΔT_b ? m Of the given solutions urea and glucose remain unionized and remain as one particle in solution. In case of Na₂SO₄(a strong electrolyte) there are three particles for each molecule(2 Na+and one \(SO_{4}^{2-}\))whereas for KNO₃(strong electrolyte) there are 2 particles (one K+ and one \(NO_{3}^{-}\)). Thus for equimolar solutions of these two, the number of particles is more in Na₂SO₄(aq) and so is molality. Thus it will exhibit highest boiling point.

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