CHEMISTRY WITH M.R.

KHAN
TOPIC- Mole Concept (Concentration Terms)
1. In one molal solution that contains 0.5 mole of 11. What is the molarity of H 2 SO4 solution, that
a solute, there is : [NEET 2022] has a density 1.84 g/cc at 350C and contains
(1) 500 mL of solvent (2) 500 g of solvent 98% by weight :
(3) 100 mL of solvent (4) 1000 g of solvent (1) 18.4 M (2) 18 M
2. The density of 2 M aqueous solution of NaOH (3) 4.18 M (4) 8.14 M
is 1.28 g/cm3 .The molality of the solution is 12. The concentration unit, independent of
[Given that molecular mass of NaOH = 40 g temperature, would be :
mol-1 ] [NEET 2019] (1) Normality (2) Weight volume percent
(1) 0.01 M (2) 0.001 M (3) Molality (4) Molarity
(3) 0.1 M (4) 0.02M 13. How many grams of CH3OH should be added
3. Which of the following is dependent on to water to prepare 150 mL solution of 2 M
temperature : [NEET 2017] CH 3 OH:
(1) Molarity (2) Mole fraction (1) 9.6  103 (2) 2.4  103
(3) Weight percentage (4) Molality
(3) 9.6 (4) 2.4
4. What is the mole fraction of the solute in a 1.00
14. Molarity of liquid HCl, if density of solution is
m aqueous solution : [NEET 2015]
1.17 g/cc is :
(1) 1.770 (2) 0.0354
(1) 36.5 (2) 18.25
(3) 0.0177 (4) 0.177
(3) 32.05 (4) 42.10
5. How many grams of concentrated nitric acid
15. 25.3 g of sodium carbonate,Na 2 CO 3 is
solution should be used to prepare 250 mL of
dissolved in enough water to make 250 mL of
2.0 M HNO 3 ?The concentrated acid is 70%
solution.If sodium carbonate dissociates
HNO 3 . [NEET 2013]
completely, molar concentration of sodium ion,
(1) 70.0 g conc.HNO3
(2) 54.0 g conc.HNO3 Na+ and carbonate ions, CO 23  are respectively
(3) 45.0 g conc.HNO3 (Molar mass of Na2CO3 = 106 g mol-1 )
(4) 90.0 g conc.HNO3 (1) 0.955 M and 1.910 M
6. 6.02 × 1020 molecules of urea are present in (2) 1.910 M and 0.955 M
100 mL of its solution. The concentration of (3) 1.90 M and 1.910 M
solution is: [NEET 2013] (4)0.477 M and 0.477 M
(1) 0.01 M (2) 0.001 M 16. Number of HCl molecules present in 10 ml of
(3) 0.1 M (4) 0.02M 0.1 N solution is:
7. Concentrated aqueous sulphuric acid is 98% (1) 6.022 × 1023
H2 SO 4 by mass and has a density of 1.80 g (2) 6.023 × 1022
mL-1.Volume of ac id reduired to make one litre (3) 6.022 × 1021
of 0.1 M H2SO4 solution is :
(4) 6.022 × 1020
(1) 16.65 mL (2) 22.20 mL
17. The total number of ions present in 1 ml of 0.1
(3) 5.55 mL (4) 11.10 mL
M barium nitrate solution is:
8. The mole fraction of the solute in one molal (1) 6.02 × 1018
aqueous solution is : (2) 6.02 × 1019
(1) 0.009 (2) 0.018 (3) 3.0 × 6.02 × 1019
(3) 0.027 (4) 11.10 mL (4) 3.0 × 6.02 × 1018
9. 2.5 litre of 1 M NaOH solution is mixed with 18. The molarity of a solution obtained by mixing
another 3 litre of 0.5 M NaOH solution.Then 750 mL of 0.5(M) HCl with 250 mL of 2(M)HCl
find out molarity of resultant solution: will be: [JEE-Main-2013]
(1) 0.80 M (2) 1.0 M (1) 1.75 M (2) 0.975 M
(3) 0.73 M (4) 0.50 M (3) 0.875 M (4) 1.00 M
10. How many gram of dibasic acid (mol.weight 200) 19. 100 ml of 0.1 M H2SO4 is mixed with 100 ml of
should be present in 100 mL of the aqueous 0.1 NaOH. The normality of the solution
solution to give strength of 0.1 N? obtained is:
(1) 10 g (2) 2 g (1) 0.4 N (2) 0.05 N
(3) 1 g (4) 20 g (3) 0.04 N (4) 0.2 N
20. What will be the molarity of a solution, which 28. Which of the following concentration term is/
contains 5.85 g of NaCl(s) per 500 mL? are temperature independent:
(1) 4 mol L–1 (1) molality (2) ppm
(2) 20 mol L–1 (3) mole fraction (4) all of these
(3) 0.2 mol L–1 29. Mole fraction of ethylene glycol [C2H6O2] in a
(4) 2 mol L–1 solution containing 20% of C2H6O2 by mass will
21. If 500 mL of a 5M solution is diluted to 1500 be:
mL, what will be the molarity of the solution (1) 0.086
obtained? (2) 0.068
(1) 1.5 M (2) 1.66 M (3) 0.932
(3) 0.017 M (4) 1.56 M (4) None
22. What will be the molality of the solution 30. 0.5 M H2SO4 is diluted from 1 litre to 10 litre,
containing 18.25 g of HCl gas in 500 g of water? normality of the resulting solution is:
(1) 0.1 m (2) 1 M (1) 1 N (2) 0.1 N
(3) 0.5 m (4) 1 m (3) 10 N (4) 11 N
23. The number of molecules in 100 ml of 0.02 N 31. Molar soluton means 1 mole of solute present
H2 SO4 is: in:
(1) 6.02 × 1022 (2) 6.02 × 1021 (1) 1000 g of solvent
20
(3) 6.02 × 10 (4) 6.02 × 1018 (2) 1 litre of solvent
24. Density of a 2.05 M solution of acetic acid (3) 1 litre of solution
inwater is 1.02 g/mL. The molality of the (4) 1000 g of solution
solution is: 32. The molarity of a solution containing 50 g of
NaCl in 500 g of a solution and having a density
(1) 1.14 mol kg–1
of 0.936 g/cm3 is:
(2) 3.28 mol kg–1
(1) 1.5 M (2) 1.6 M
(3) 2.28 mol kg–1 (3) 1.8 M (4) 1.2 M
(4) 0.44 mol kg–1 33. 20 mL of 0.5 M HCl is mixed with 30 mL of 0.3
25. The mole fraction of a given sample of I2 in C6H6 M HCl, the molarity of the resulting solution
is 0.2. The molality of I2 in C6H6 is: is:
(1) 0.32 (1) 0.8 M (2) 0.53 M
(3) 0.38 M (4) 0.83 M
(2) 3.2
34. How may moles and how many grams of NaCl
(3) 0.032
are present in 250 mL of a 0.5 M NaCl solution?
(4) 0.48 (1) 0.125 mol; 7.32 g
26. Concentrated aqueous sulphuric acid 98% (2) 7.32 mol; 0.125 g
H2 SO 4 by mass has a density of 1.80 g mL – (3) 0.125 mol; 0.125 g
1
Volume of acid required to make one litre of (4) 7.32 mol; 7.32 g
0.1 M H2SO4 solution is: 35. The density (in g mL –1 ) of a 3.6 M sulphuric
(1) 16.65 mL acid solution, i.e., 29% H 2SO4 (molar mass =
98 g mol–1) by mass will be:
(2) 22.20 mL
(1) 1.45
(3) 5.55 mL (2) 1.64
(4) 11.10 mL (3) 1.88
27. Molarity of H2SO4 (density 1.8 g/mL) is 18 M. (4) 1.22
The molality of this H2SO4 is : 36. 1 litre solution containing 490 g of sulphuric
(1) 36 acid is diluted to 10 litre with water. What is
(2) 200 the normality of the resulting solution?
(3) 500 (1) 0.5 N (2) 1.0 N
(4) 18 (3) 5.0 N (4) 10.0 N