If 2.74 g of the metal oxide contains 1.53 g of metal, then the empirical formula of vanadium oxide is:

(Atomic Mass of V = 52)

1. ${\mathrm{V}}_2{\mathrm{O}}_3$

2. $\mathrm{VO}$

3. ${\mathrm{V}}_2{\mathrm{O}}_5$

4. ${\mathrm{V}}_2{\mathrm{O}}_7$

A gaseous mixture of propane and butane of volume 3 litre on complete combustion produces 11.0 litre $C{O}_2$ under standard conditions of temperature and pressure. The ratio of volume of butane to propane is:

1. 1 : 2

2. 2 : 1

3. 3 : 2

4. 3 : 1

The percentage by volume of $C_3{H}_8$ in a gaseous mixture of $C_3{H}_8$, $C{H}_4$ and CO is 20. When 100 mL of the mixture is burnt in excess of $O_2$, the volume of $C{O}_2$ produced is:

1. 90 mL

2. 160 mL

3. 140 mL

4. None of the above

The percentage of oxygen present in the compound ${\mathrm{CaCO}}_3.3{\mathrm{Ca}}_3 {\left({\mathrm{PO}}_4\right)}_2$ is-

1.  23.3%

2.  45.36%

3.  41.94%

4.  17.08%

A gaseous compound is composed of 85.7% by mass carbon and 14.3% by mass hydrogen. It's density is 2.28 g/litre at 300 K and 1.0 atm pressure. Determine the molecular formula of the compound:

1.  ${\mathrm{C}}_2{\mathrm{H}}_2$

2.  ${\mathrm{C}}_2{\mathrm{H}}_4$

3.  ${\mathrm{C}}_4{\mathrm{H}}_8$

4.  ${\mathrm{C}}_4{\mathrm{H}}_{10}$

Calculate the % of free ${\mathrm{SO}}_3$ in oleum (a solution of ${\mathrm{SO}}_3$ in ${\mathrm{H}}_2{\mathrm{SO}}_4$) that is labelled 109% ${\mathrm{H}}_2{\mathrm{SO}}_4$.

1. 40

2. 30

3. 50

4. None

Suppose two elements X and Y combine to form two compounds ${\mathrm{XY}}_2$ and ${\mathrm{X}}_2{\mathrm{Y}}_3$when 0.05 mole of${\mathrm{XY}}_2$ weighs 5 g while 3.011$\times$10²³ molecules of ${\mathrm{X}}_2{\mathrm{Y}}_3$ weighs 85 g. The atomic masses of x and y are respectively:

1.  20, 30

2.  30, 40

3.  40, 30

4.  80, 60

100 mL of 10% NaOH (w/V) is added to 100 mL of 10% HCl (w/V). The resultant solution becomes:

1. Alkaline

2. Strongly alkaline

3. Acidic

4. Neutral

1.44 gram of titanium (At. wt. =48) reacted with excess of ${\mathrm{O}}_2$ and produce x gram of non-stoichiometric compound ${\mathrm{Ti}}_{1.44} \mathrm{O}.$ The value of x is:

1.  2

2.  1.77

3.  1.44

4.  None of the above

For the reaction; 
\(2Fe{(NO_3)}_3 + 3Na_2CO_3 \rightarrow Fe_2{(CO3)}_3 + 6NaNO_3\)
Initially if 2.5 mole of \(Fe{(NO_3)}_3 \) and 3.6 mole of \(Na_2CO_3\) is taken. If 6.3 mole of \(NaNO_3\) is obtained then % yield of given reaction is:

1.  50

2.  84

3.  87.5

4.  100