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

ln, a closed vessel, 50 ml of $A_2{B}_3$ completely reacts with 200 ml of C₂ according to the following equation:

$2A_2{B}_3\left(g\right) + 5C_2\left(g\right)\to 3C_3{B}_2\left(g\right)+C{A}_4\left(g\right)$

The composition of the gaseous mixture in the system will be:

1. 100 ml $C_2, 50 ml C_3{B}_2,50 ml C{A}_4$

2. 25 ml $C_2, 75 ml C_3{B}_{\mathrm{2,}} 25 ml C{A}_4$

3. 75 ml $C_2, 75 ml C_3{B}_{2, }25 ml C{A}_4$

4. 10 ml $C_2, 25 ml C_3{B}_{2, }100 ml C{A}_4$

On complete combustion, 44 g of a sample of a compound gives 88 g CO₂ and 36 g of H₂O. The molecular formula of the compound may be:

1.  ${\mathrm{C}}_4{\mathrm{H}}_6$ $$

2.  ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ $$

3.  ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$

3.  ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$

To produce 20 moles of ammonia via Haber's process, how many moles of hydrogen molecules are required?

An organic compound contains carbon, hydrogen, and oxygen. Its elemental analysis gave C, 38.71%, and H, 9.67%. The empirical formula of the compound would be:

The volume of oxygen gas (O₂) needed to completely burn 1 L of propane gas (C₃H₈)  (both O₂ & propane measured at 0°C and 1 atm) will be:
1. 7 L
2. 6 L
3. 5 L
4. 10 L

25.3 g of sodium carbonate, Na₂CO₃, is dissolved in enough water to make 250 mL of a solution. If sodium carbonate completely dissociates, the molar concentrations of sodium ion, Na⁺ and carbonate ion, $C{O}_3^{2-}$are respectively:
(molar mass of Na₂CO₃ = 106g mol^-1)

1. 0.955 M and 1.910 M
2. 1.910 M and 0.955 M
3. 1.90 M and 1.910 M
4. 0.477 M and 0.477 M

 The maximum number of molecules is present in which of the following? 

1. 15 L of $H_2$ gas at STP

2. 5 L of $N_2$ gas at STP

3. 0.5 g of $H_2$ gas

4. 10 g of $O_2$ gas

The number of  moles  present in 2.5 litres of 0.2 M  ${\mathrm{H}}_2{\mathrm{SO}}_4$ are:

1. 0.25

2. 0.5

3. 0.75

4. 0.2

The number of valence electrons present in 0.53 grams of Na₂CO₃ is:

1. $3.01\times {10}^{23}$

2. $7.22\times {10}^{22}$

3. $12.046\times {10}^{23}$

4. $6.023\times {10}^{23}$