The mole fraction of the solute in a 1.00 molal aqueous solution is:

 1 0.00177 2 0.0344 3 0.0177 4 0.177
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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

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The density of a 2 M aqueous solution of NaOH is 1.28 g/$c{m}^{3}$. The molality of the solution is:
[molecular mass of NaOH = 40 $gmo{l}^{-1}$]

 1 1.20 m 2 1.56 m 3 1.67 m 4 1.32 m
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Concentrated nitric acid is 70% HNO3. The amount of concentrated nitric acid solution that should be used to prepare 250 mL of 2.0 M HNO3 would be:

1. 90.0 g conc. HNO3
2. 70.0 g conc. HNO3
3. 54.0 g conc. HNO3
4. 45.0 g conc. HNO3

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A solution is prepared by adding 2 g of substance A to 18 g of water. The mass percent of the solute is-

1. 20%

2. 10%

3. 15%

4. 18%

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An aqueous solution of urea containing 18 g of urea in 1500 cm3 of the solution has a density of 1.052 g/cm3. If the molecular weight of urea is 60, then the molality of the solution is-
1. 0.2
2. 0.192
3. 0.064
4. 1.2

Subtopic:  Concentration Based Problem |
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Sulphuric acid reacts with sodium hydroxide as follows

${\mathrm{H}}_{2}{\mathrm{SO}}_{4}+2\mathrm{NaOH}\to {\mathrm{Na}}_{2}{\mathrm{SO}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}$

When 1L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained are respectively-

1. 0.1 M, 7.10 g

2. 7.10 g, 0.025 M

3. 0.025 M, 3.55 g

4. 3.55 g, 0.25 M

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The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na+ per ml is:

(Rounded off to the nearest integer) [Given: Atomic weight in g mol–1 – Na: 23; N: 14; O: 16]

1. 13 g

2. 26 g

3. 18 g

4. 22 g

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