The concentration of 69% nitric acid (by mass) in a sample with a density of 1.41 g / mL will be-

1. 15.44 mol/L

2. 17.14 mol/L

3. 20.06 mol/L

4. 12.26 mol/L

Subtopic:  Concentration Based Problem |
64%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

20 g of sugar is dissolved in enough water to make a final volume of 2L. The concentration of sugar (C12H22O11) in mol L–1  will be:

1. 0.29 mol/L

2. 0.029 mol/L

3. 0.35 mol/L

4. 0.032 mol/L

Subtopic:  Concentration Based Problem |
77%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

If the density of methanol is 0.793 kg L–1, the volume needed for making 2.5 L of its 0.25 M solution would be:

1. 22.25 mL

2. 24.78 mL

3. 25.22 mL

4. 22.52 mL

Subtopic:  Concentration Based Problem |
60%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

The chloroform contamination level in water is 15 ppm (by mass of chloroform). The molality of chloroform in the water sample would be:

1. 3.25 × 10-4

2. 1.5 × 10-3

3.  7.5 × 10-3

4. 1.25 × 10-4

Subtopic:  Concentration Based Problem |
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

The molarity of a solution of ethanol in water having a mole fraction of ethanol is 0.040 (assume the density of water to be 1) would be -

1. 2.143 M

2. 2.314 M

3. 2.413 M

4. 2.141 M

Subtopic:  Concentration Based Problem |
62%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch

The mass of CaCO3 required to react completely with 25 mL of 0.75 M HCl according to the given reaction would be:

CaCO3(s) + HCl(aq) ➡ CaCl2(aq) + CO2(g) + H2O(l)

1. 0.36 g

2. 0.09 g

3. 0.96 g

4. 0.66 g

Subtopic:  Concentration Based Problem | Equation Based Problem |
62%
From NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints