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

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

CaCO_3(s) + HCl_(aq) ➡ CaCl_2(aq) + CO_2(g) + H₂O_(l) 

1. 0.36 g

2. 0.09 g

3. 0.96 g

4. 0.66 g

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 $gmo{l}^{-1}$]

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

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%

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

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

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