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# Solutions

#### Multiple Choice Questions

1.

18 g glucose (C6H12O6 ) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:

• 76.0

• 752.4

• 759.0

• 7.6

B.

752.4

Vapour pressure of water (p°) = 760 torr

Number of moles of glucose

Molar mass of water = 18 g/mol
Mass of water (given) = 178.2 g

Number of moles of water

Total number of moles = (0.1 +9.9) = 10 moles

Now, the mole fraction of glucose in solution = Change in pressure with respect to initial pressure.

therefore, Vapour pressure of solution = (760-7.6)torr
=752.4 torr

422 Views

2.

Kf for water is 1.86K kg mol–1.If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8°C?

• 72 g

• 93 g

• 39 g

• 27 g

B.

93 g

Coolant is glycol (C2H6O2) and is non-electrolyte.
ΔTf =2.8°

314 Views

3.

The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL. The molarity of this solution is

• 0.50 M

• 1.78 M

• 1.02 M

• 2.05

D.

2.05

Total mass of solution
= 1000 g water +120 g urea
= 1120 g
283 Views

4.

The ratio of masses of oxygen and nitrogen of a particular gaseous mixture is 1:4. The ratio of number of their molecule is

• 1:4

• 7:32

• 1:8

• 3:16

B.

7:32

231 Views

5.

For the estimation of nitrogen 1.4g of an organic compound was digested by Kjeldahl's method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 mL of M/10 sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is

• 6%

• 10%

• 3%

• 5%

B.

10%

386 Views

6.

A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution?

• 0.100

• 0.190

• 0.086

• 0.050

C.

0.086

596 Views

7.

Consider separate solution of 0.500 M C2H5OH (aq), 0.100 M Mg3(PO4)2 (aq) 0.250 M KBr (aq) and 0.125 M Na3PO4 (aq) at 25oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

• They all have the same osmotic pressure

• 0.100 M Mg3(PO4)2 (aq) has the highest osmotic pressure

• 0.125 M Na3PO (aq) has the highest osmotic pressure

• 0.500 M C2H5OH (aq) has the higest osmotic pressure

A.

They all have the same osmotic pressure

effective molarity = Van't Hoff factor x molarity
0.5 M C2H5OH (aq)      i =1
Effective molarity = 0.5

0.25 M KBr (aq)          i = 2
Effective molarity  = 0.5

0.1 M Mg3(PO4)2 (aq) i = 5
Effective molarity = 0.5 M

0.125 M Na3PO4 (aq)
Effective molarity = 0.5 M

Hence, all colligative properties are same

960 Views

8.

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

• 18 mg

• 36 mg

• 42 mg

• 54 mg

A.

18 mg

The initial strength of acetic acid = 0.06N
Final strength = 0.042 N
Volume given = 50 mL
there Initial m moles of CH3COOH
= 0.06 x 50 = 3
Final m moles of CH3COOH
= 0.042 x 50 = 21
therefore, m moles of CH3COOH absorbed
= 3-2.1
= 0.9 m mol
Hence, mass of CH3COOH  absorbed per gram of charcoal
=

424 Views

9.

The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1 ) of the substance is:

• 32

• 64

• 128

• 488

B.

64

Given,
po = 185 torr at 20oC
ps = 183 torr at 20oC
Mass of non-volatile substance,
m= 1.2 g
Mass of acetone taken = 100 g
As we have,
putting the values, we get

514 Views

10.

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be

• 0.875 M

• 1.00 M

• 1.75 M

• 0.0975 M

A.

0.875 M

M1V1 + M2V2 = MV

M= 0.875
215 Views