jyoti

Al2 (SO4)3  {d= 1.253 gm/mL } contain 22% salt by weight. The molarity, normality and molality of the solution is a. 0.805 M, 4. 83 N, O.825 m b. 0.805 M, 48.3 N, O.80s m c. 4.83 M, 4. 83 …

a. Molarity b. Molality c. Formality d. Normality Ans. b Explanation :  Calculation of molality does not have a volume  component. (16)

mol-1   is dissolved in 100 mL of water to obtain a solution whose density is 1.25 g mL-1  The molarity and molality of the solution are respectively a. 0.75 and 1 b. 0.8 and 1 c. 1 and 0.8 d. 1 …

a. 1m b. 0.5 m c. 0.2 m d. 2m Ans. c (6)

a. ppm by weight of MgSO4 b. g/L of CaCO3  and MgCO3 c. ppm by weight of CaCO 3 irrespective of whether it is actually present d. ppm of CaCO3  actully present in water Ans. c (8)

a.  14 b. 3.2 c. 4 d. 2 Ans. b (8)

a. 0.027 b. 0.036 c. 0.018 d. 0.009 Ans. c (9)

a. 250 b. 200 c. 400 d. 1000 Ans. c (6)

a. Molality b. Weight fraciton of solute c. Fraction of solute present in water d. Mole fraction Ans. c Explanation : Since fraction of solute present in water depends on the volume of water, it will change with the increase …

a. o.177 b. 1.770 c. 0.0354 d. 0.0177 Ans. d (5)

a. 0.1 b. 0.9 c. 0.3 d. 0.6 Ans. d (16)

H2 O. The mole fraciton of urea is a. 10/10.1 b. 10.1/10 c.10.1/0.1 d.  0.1/10.1 Ans. d (9)

a. 1000 g of the solvent b. One litre of the solvent c. One litre of the solution d. 22.4 litres of the solution Ans. a Explanation : The number of moles of solute dissolved in 1000 g of the …

CaCO3  contains 10 gms of carbonate. Concentration of the solution is a. 10 ppm b. 100 ppm c. 1000 ppm d. 10000 ppm Ans. d (10)

H2  O2  is a. 77/68 b. 68/77 c. 20/80 d. 80/20 Ans. b (39)

a. Gram per million b. Milligram percent c. Microgram percent d. Nano gram percent e. Part per million Ans. e (11)

a.  0. 176 b. 3. 52 c. 1. 78 d. 0.88 e. 17. 8 Ans. c (8)

H2 SO4 and 30 mL of N/3 HNO3  are mixed together and volume made to one litre. The normality of the resulting solution is a. N/5 b. N/10 c. N/20 d. N/40 e. N/ 25 Ans. d (6)

H 2 SO4  has a density of 1.98 g/mL and is 98% H2 SO4 by weight. Its normality is a. 2N b. 19.8 N c. 39.6 N d. 98 N Ans. c (6)

0.2 N – NaOH solution. a. 0.45 g b. 0.90 g c. 1.08 g d. 1. 26 g Ans. d (38)

a.  4.0N H2 SO4 b. 0.5 N H2 SO4 c. 1 N H2 SO4 d. 2N H2  SO4 Ans. a (13)

cm3  of   0.1 N monobasic acid requires 15 cm3   of sodium hydroxide solution whose normality is a. 0.066 N b. 0. 66 N c. 1.5 N d. 0. 15 N Ans. a (4)

20°  C and NaCl concentration is 3.5% the normality of the sea water is a.  0.65 N b. 0.68 N c. 0.66 N d. 0.61 N Answer : d (4)

a.  o.97 N b. 0.142 N c. 0.194 N d. 0.244 N Ans. c (12)

N H2 SO4  and the mixture is made one litre. Normality of the mixture will be a.  0.56 N b. 0.50 N c. 0.40 N d. 0.35 N Ans. b (6)

a. 13. 25 g b. 26. 5 g c. 53 g d. 6.125 g Ans. a (10)

H2 SO4  (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could be a. 0.18 N b. 0.09 N c. 0.36 N d. 1800 N Ans. c (12)

a.  HCl b. perchloric acid c. HNO3 d. phosphoric acid Ans. d (10)

a. 40 mL b. 20mL c. 10mL d. 60 mL Ans. a (8)

a. 2N b.4N c. N/2 d. N/4 Ans. b Explanation : N = M x basicity ; N = 2 x 2 = 4 (14)

a. 0. 5 L of  A + 1.5 L of B b. 1.5 L of  A + 0.5 L of B c. 1.0 L of  A + 1.0 L of B d. 0. 75 L of  A + 1.25 L …

1.2046 x 106  hydrochloric acid molecules in one dm3  of the solution . The strength of the solution is a. 6 N b. 2N c. 4N d. 8 N Ans. b (13)

10N HNO3  to get  0.1 N HNO3  a. 1000 mL b. 990 mL c. 1010 mL d. 10 mL Ans. b Explanation : N1  V1  =  N2  V2 10 x 10 = 0.1 (10+V) V = 10 x 10 / …