मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL−1.

Advertisements
Advertisements

प्रश्न

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL−1.

संख्यात्मक
Advertisements

उत्तर

Given: Let mass of solution in water = 100 g, then mass of KI = 20 g

∴ Mass of solvent (water) = 100 − 20 = 80 g = 0.080 kg

(a) Molar mass of KI = 39 + 127 = 166 g mol−1

∴ Moles of KI = `(20  "g")/(166  "g mol"^(-1))`

= 0.120 mol

Molality of solution = `"Number of moles of KI"/"Mass of solvent in kg"`

= `(0.120  "mol")/(0.080  "kg")`

= 1.5 mol kg−1

(b) Density of the solution = 1.202 g mL−1

∴ Volume of solution = `"Mass"/"Density"`

= `(100  "g")/(1.202  "g mL"^(-1))`

= 83.2 mL

= 0.0832 L

Molarity of the solution = `"Number of moles of solute"/"Volume of solution in L"`

= `(0.120  "mol")/(0.0832  "L")`

= 1.44 M

(c) No. of moles of KI = 0.120 mol

No. of moles of water = `"Mass of water"/"Molar mass of water"`

= `(80  "g")/(18  "g mol"^(-1))`

= 4.44 mol

Mole fraction of KI = `"Number of moles of KI"/"Total number of moles in solution"`

= `0.12/(0.120 + 4.44)`

= `0.120/4.560`

= 0.0263

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Solutions - Intext Questions [पृष्ठ ५]

APPEARS IN

एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 1 Solutions
Intext Questions | Q 1.5 | पृष्ठ ५

संबंधित प्रश्‍न

Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.


Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL−1?


An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL−1, then what shall be the molarity of the solution?


Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.


22.22 gram of urea was dissolved in 300 grams of water. Calculate the number of moles of urea and molality of the urea solution.

(Given: Molar mass of urea = 60 gram mol−1)


A sample of 12 M Concentrated hydrochloric acid has a density 1.2 M gL-1 calculate the molality.


1 M, 2.5 litre NaOH solution is mixed with another 0.5 M, 3 litre NaOH solution. Then find out the molarity of the resultant solution:


An X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is ____________.


25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 ml. The molarity of barium hydroxide solution was ______.


Which of the following concentration unit is independent of temperature?


Out of molality (m), molarity (M), formality (F) and mole fraction (x), those which are independent of temperature are:


Match the terms given in Column I with expressions given in Column II.

Column I Column II
(i) Mass percentage  (a) `"Number of moles of the solute component"/"Volume of solution in litres"`
(ii) Volume percentage  (b) `"Number of moles of a component"/"Total number of moles of all the components"`
(iii) Mole fraction (c) `"Volume of the solute component in solution"/"Total volume of solution" xx 100`
(iv) Molality (d) `"Mass of the solute component in solution"/"Total mass of the solution" xx 100`
(v) Molarity (e) `"Number of moles of the solute components"/"Mass of solvent in kilograms"`

What is the ratio of mass of an electron to the mass of a proton?


The number of electrons involved in the reduction of one nitrate ion to hydrazine is


What is the normality of 0.3 m H3Pcl solution?


250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ______ × 1021. (Nearest integer) (NA = 6.022 × 1023).


A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm−3. The molarity of the solution is ______ mol dm−3. (Round off to the Nearest Integer)

[Atomic masses: K: 39.0 u; O: 16.0 u; H: 1.0 u]


The mole fraction of a solute in a 100 molal aqueous solution is ______ × 10-2. (Round off to the Nearest Integer).

[Given :Atomic masses : H : 1.0 u, O : 16.0 u]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×