The volume of oxygen at STP required to completely burn 30 ml of acetylene at STP is [Orissa JEE 1997]

(1) 100 ml

(2) 75 ml

(3) 50 ml

(4) 25 ml

If Avogadro number N_A, is changed from 6.022 x 10²³ mol^-1 to 6.022 x 10²⁰ mol^-1 this would change

(1) the definition of mass in units of grams

(2) the mass of one mole of carbon

(3) the ratio of chemical species to each other in a balanced equation

(4) the ratio of elements to each other in a compound

6.02 x 10²⁰ molecules of urea are present in 100 mL of its solution. The concentration of the solution is

(a) 0.02 M

(b) 0.01 M

(c) 0.001 M

(d) 0.1 M

A mixture of methane and ethene in the molar ratio of x : y has a mean molar mass of 20. What would be the mean molar mass. if the gases are mixed in the molar ratio of y : x?

(1) 22                               

(2) 24

(3) 20.8                           

(4) 19

The volume of a drop of water is 0.0018 mL then the number of water molecules present in two drop of water at room temperature is:

1.  $12.046\times {10}^{19}$

2.  $1.084\times {10}^{18}$

3.  $4.84\times {10}^{17}$

4.  $6.023\times {10}^{23}$

Carbon and oxygen combine to form two oxides, carbon monoxide and carbon dioxide in which the ratio of the weights of carbon and oxygen is respectively 12 : 16 and 12 : 32. These figures illustrate the:

1.  Law of multiple proportions

2.  Law of reciprocal proportions

3.  Law of conservation of mass

4.  Law of constant proportions

The formula of an acid is ${\mathrm{HXO}}_2.$ The mass of 0.0242 moles of the acid 1.657 g. What is the atomic weight of X?

1.  35.5

2.  28.1

3.  128

4.  19.0

A 6.85 g sample of the hydrates $\mathrm{Sr}{\left(\mathrm{OH}\right)}_2.{\mathrm{xH}}_2\mathrm{O}$ is dried in an oven to give 3.13 g of anhydrous $\mathrm{Sr}{\left(\mathrm{OH}\right)}_2$. What is the value of x?   (Atomic weights : Sr=87.60, O=16.0, H=1.0)

1.  8

2.  12

3.  10

4.  6

A sample of phosphorus that weights 12.4 g exerts a pressure 8 atm in a 0.821 litre closed vessel at 527$°$C. The molecular formula of the phosphorus vapour is:

1.  ${\mathrm{P}}_2$

2.  ${\mathrm{P}}_4$

3.  ${\mathrm{P}}_6$

4.  ${\mathrm{P}}_8$

Phosphoric acid $\left({\mathrm{H}}_3{\mathrm{PO}}_4\right)$ prepared in a two step process.

(1) ${\mathrm{P}}_4+5{\mathrm{O}}_2 \stackrel{}{\to } {\mathrm{P}}_4{\mathrm{O}}_{10}$
(2) ${\mathrm{P}}_4{\mathrm{O}}_{10}+6{\mathrm{H}}_2\mathrm{O} \stackrel{}{\to } 4{\mathrm{H}}_3{\mathrm{PO}}_4$

We allow 62g of phosphorus to react with react with excess oxygen which form ${\mathrm{P}}_4{\mathrm{O}}_{10}$ in 85% yield. In the step (2) reaction 90% yield of ${\mathrm{H}}_3{\mathrm{PO}}_4$ is obtained. Produced mass of ${\mathrm{H}}_3{\mathrm{PO}}_4$ is:

1.  37.485 g

2.  149.949 g

3.  125.47 g

4.  564.48 g