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Solutions Quiz-4

Solutions Quiz-4

Difficulty Level: Easy
10 Questions
3 Plays
QuizardQuizard

Popular Questions InSolutions Quiz-4

1

A solution containing 1.8g of a compound (empirical formula CH2OCH_2OCH2​O) in 40g of water is observed to freeze at −0.465∘C-0.465^\circ C−0.465∘C. The molecular formula of the compound is (Kf=1.86 kg K mol−1K_f = 1.86\text{ kg K mol}^{-1}Kf​=1.86 kg K mol−1)

2

KfK_fKf​ for water is 1.86 K kg mol−11.86\text{ K kg mol}^{-1}1.86 K kg mol−1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2C_2H_6O_2C2​H6​O2​) must you add to get the freezing point of the solution lowered by 2.8∘C2.8^\circ C2.8∘C?

3

Calculate the molal depression constant of a solvent which has freezing point 16.6∘C16.6^\circ C16.6∘C and latent heat of fusion 180.75 J g−1.180.75\, J\, g^{-1}.180.75Jg−1..

4

The freezing point of water is depressed by 0.37∘C0.37^\circ C0.37∘C in a 0.01 molal NaCl solution. The freezing point of 0.02 molal solution of urea is depressed by.

5

The freezing point depression constant of water is 1.86∘C m1.86^\circ C\,m1.86∘Cm. If 5.00 g of Na2SO4Na_2SO_4Na2​SO4​ is dissolved in 45.0g of H2OH_2OH2​O, the freezing point is found to be −3.82∘C-3.82^\circ C−3.82∘C. Calculate the can’t Hoff factor for Na2SO4Na_2SO_4Na2​SO4​.

6

0.01 M solution of KCl and BaCl2BaCl_2BaCl2​ are prepared in water. The freezing point of KCl solution is fount to be−2∘C-2^\circ C−2∘C. What is the freezing point of BaCl2BaCl_2BaCl2​ solution, assuming complete Ionization?

7

Van't Hoff factors of aqueous solution of X, Y and Z are 1.8, 0.8 and 2.5 respectively. Hence, correct order of the colligative property will be?

8

If osmotic pressure of 1 M urea is π\piπ, what will be the osmotic pressure for 0.1 M NaCl?

9

An aqueous solution containing 1.248g of barium chloride (molar mass = 208.34g mol−1mol^{-1}mol−1) in 100g of water boils at 100.0832∘C100.0832^{\circ} C100.0832∘C. Calculate the degree of dissociation of BaCl2BaCl_2BaCl2​ (KfK_fKf​ for water=0.52 K kg mol−1= 0.52\text{ K kg mol}^{-1}=0.52 K kg mol−1)

10

The Henry’s law constant for the solubility of N2N_2N2​ gas in water at 298 K is 1.0×1051.0 \times 10^{5}1.0×105 atm. The mole fraction of N2N_2N2​ in air is 0.8. The number of moles of N2N_2N2​ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is.

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