English
Karnataka Board PUCPUC Science 2nd PUC Class 12

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

Advertisements
Advertisements

Question

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?

Numerical
Advertisements

Solution 1

Molar mass of ethylene glycol [C2H4(OH)2] = 2 × 12 + 6 × 1 + 2 × 16

= 62 g mol−1

Number of moles of ethylene glycol = `(222.6  g)/(62  g  "mol"^(-1))`

= 3.59 mol

∴ Molality of the solution = `(3.59  "mol")/(0.200  kg)`

= 17.95 m

Total mass of the solution = (222.6 + 200) g

= 422.6 g

Density of the solution = 1.072 g mL−1    ...[Given]

∴ Volume of the solution = `(422.6  g)/(1.072  g  mL^(-1))`

= 394.2 mL

= 0.3942 L

⇒ Molarity of the solution = `(3.59  mol)/(0.3942  L)`

= 9.11 M

shaalaa.com

Solution 2

Molality (m) of the solution is given by:

m = `w/(M') xx 1000/(w')`

In the present case,

w = 222.6 g,

M' = 62 g mol−1

w' = 200 g

∴ m = `(w xx 1000)/(M' xx w')`

= `(222.6 xx 1000)/(62 xx 200)`

= 17.95 mol kg−1

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

= `(222.6 + 200)/(1.072)`

= 394.2 mL

Molarity is given by:

w = `(M xx M' xx v)/1000`

∴ `M = (w xx 1000)/(M' xx v)`

= `(222.6 xx 1000)/(62 xx 394.2)`

= 9.11 mol L−1

shaalaa.com
  Is there an error in this question or solution?
Chapter 1: Solutions - Exercises [Page 28]

APPEARS IN

NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 1 Solutions
Exercises | Q 1.8 | Page 28
Nootan Chemistry [English] Class 12 ISC
Chapter 1 Solutions
'NCERT TEXT-BOOK' Exercises | Q 2.8 | Page 124

RELATED QUESTIONS

Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.


Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.


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)


When a solute is present in trace quantities the following expression is used:


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 ____________.


The molarity of pure water is ____________.


An aqueous solution of glucose is 10% in strength. The volume in which 1 g mole of it is dissolved, will be:


2.5 litres of NaCl solution contain 5 moles of the solute. What is the molarity?


The mole fraction of the solute in one molal aqueous solution is ____________.


What is the normality of a 1 M solution of H3PO4?


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


Which of the following is a correct statement for C2H5Br?


Given below are two statements labelled as Assertion (A) and Reason (R).

Assertion (A): Molarity of a solution changes with temperature.

Reason (R): Molarity is a colligative property.

Select the most appropriate answer from the options given below:


Calculated the mole fraction of benzene in a solution containing 30% by mass of its is carbon tetrachloride


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 given solution of H2O2 is 30 volumes. Its concentration in terms of molarity is ______.


The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×