Advertisements
Advertisements
Question
An organic substance (M = 169 gram mol–1) is dissolved in 2000 cm3 of water. Its osmotic pressure at 12°C was found to be 0.54 atm. If R = 0.0821 L atm K–1 mol–1, calculate the mass of the solute.
Solution
Given: T = 273.15 + 12 = 285.15 K
V = 2000 cm3 = 2 L
M2= 169 g mol–1
R = 0.0821 L atm K–1 mol–1
π = 0.54 atm
To find: Mass of solute (W2)
`pi=(W_2RT)/(M2V)`
`W_2=(piM_2V)/(RT)`
`W_2=(0.54xx169xx2)/(0.0821xx285.15)`
`W_2=182.52/23.4108=7.796 g`
APPEARS IN
RELATED QUESTIONS
The vapour pressure of pure benzene is 640mm og Hg. 2.175×10-3kg of non-vloatile solute is added to 39 gram of benzene the vapour pressure of solution is 600mm of HG. Calculate molar mass of solute.
[C = 12, H = 1]
The determination of molar mass from elevation in boiling point is called as
- cryoscopy
- colorimetry
- ebullioscopy
- spectroscopy
Colligative property depends only on ........................ in a solution.
A solution of glucose in water is labelled as 10% (W/W).
Calculate:
a. Molality
b. Molarity of the solution.
[Given: Density of solution is 1.20 g mL-1 and molar mass of glucose is 180 g mol-1 ]
Calculate the mole fraction and molality of HNO3 in solution contaning 12.2%HNO3 (Given atomic mases:H=1, N=13,O=16)
Calculate the amount of CaCl2 (van't Hoff factor i = 2.47) dissolved in 2.5 L solution so that its osmotic pressure at 300K is 0.75 atmosphere.
Given : Molar mass of CaCl2 is 111g mol-1
R=0.082 L.atm K-1mol-1
A solution of glucose (molar mass = 180 g mol–1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution ? (Density of solution = 1.2 g mL–1)
A solution of a substance having mass 1.8 x 10-3 kg has the osmotic pressure of 0.52 atm at 280 K. Calculate the molar mass of the substance used.
[Volume = 1 dm3, R = 8.314 J K-1 mol-1]
The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute.
[Kb for benzene = 2.53 K kg mol-1]
Write one difference in Coagulation and Peptization
Out of 0.1 molal aqueous solution of glucose and 0.1 molal aqueous solution of KCl, which one will have higher boiling point and why?
A solution 0.1 M of Na2SO4 is dissolved to the extent of 95%. What would be its osmotic pressure at 027 ? C (R = 0.0821 L atm K-1 mol-1) .
Define the following terms :
a. Cryoscopic constant
b. Resistivity
4.0 grams of NaOH ( Molar mass = 40.0 g mol-1 ) is dissolved in 500 cm3 of water. What is the molarity of NaOH solution?
The colligative property is not represented by ____________.
Colligative properties depend on ______.
Colligative properties are observed when:
(i) a non volatile solid is dissolved in a volatile liquid.
(ii) a non volatile liquid is dissolved in another volatile liquid.
(iii) a gas is dissolved in non volatile liquid.
(iv) a volatile liquid is dissolved in another volatile liquid.
Which of the following is a colligative property?
Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K The depression in freezing point of 4.5 molal solution of glucose in water is 4 K. The ratio of molal elevation constant to molal depression constant (Kb/Kf) is ______.
