The molarity of a solution obtained by mixing 750 mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be :

0.875 M
1.00 M
1.75 M
0.975 M

Solution

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution,the change in freezing point of water (ΔT_f), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is(K_f = 1.86K kg mol^-1)

0.372K
0.0558 K
0.0744 K
0.0186 K

Solution

A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution?

The solution is non-ideal, showing – ve deviation from Raoult’s Law.
The solution is non-ideal, showing + ve deviation from Raoult’s Law
n-heptane shows + ve deviation while ethanol shows– ve deviation from Raoult’s Law.
The solution formed is an ideal solution.

Solution

For this solution inter molecular interactions between n-heptane and ethanol are weaker than n-heptane-n-heptane & ethanol-ethanol interactions hence the solution of n-heptane and ethanol is non-ideal and shows positive deviation from Raoult’s law.

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm Hg at 300 K. The vapour pressure of propyl alcohol is 200 mm Hg. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm Hg) at the same temperature will be

Solution

PA and PB are the vapour pressure of pure liquid components,A and B, respectively of an ideal binary solution. If X_A represents the mole fraction of component A, the total pressure of the solution will be.

P_A + X_A(P_B – P_A)
P_A + X_A(P_A – P_B)
P_B + X_A (P_B – P_A)
P_B + X_A(P_A – P_B)

Solution

200 mL of an aqueous solution of a protein contains its 1.26 g.The osmotic pressure of this solution at 300 K is found to be 2.57 × 10^-3 bar. The molar mass of protein will be (R = 0.083 L bar mol^-1 K^-1)

51022 g mol^-1
122044 g mol^-1
31011 g mol^-1
61038 g mol^-1

Solution

A 0.1 molal aqueous solution of a weak acid is 30% ionized.If K_f for water is 1.86°C/m, the freezing point of the solution will be :

– 0.18°C
– 0.54°C
–0.36°C
– 0.24°C

Solution

Mole fraction of the solute in a 1.00 molal aqueous solution is

0.1770
0.0177
0.0344
1.7700

Solution

The van’t Hoff factor if or a compound which undergoes dissociation in one solvent and association in other solvent is respectively :

less than one and greater than one.
less than one and less than one.
greater than one and less than one.
greater than one and greater than one

Solution

If compound dissociates in solvent i> 1 and on association i < 1.

The freezing point depression constant for water is –1.86º Cm^-1. If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by– 3.82ºC. Calculate the van’t Hoff factor for Na₂SO₄

2.05
2.62
3.11
0.381

Solution

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