Q: The volume strength of 3N\:H_{2}O_{2} solution is:

3\:N means 3 g eqwt. in 1 L solution. N=\dfrac{\text{weight}}{\text{eq. wt.}} . ( eq. wt. for hydrogen peroxide = 17 g ) Therefore, 3 = \dfrac{\text{wt}}{17} or \text{wt} = 3 \times17 = 51 g.Now, from the equation, we know that 68 g of H_2O_2 gives 22.4 L of Oxygen. Therefore, 51 g of hydrogen peroxide gives 22.4 \times \dfrac{51}{68} = 16.8 L.Thus, volume strength is 16.8 L, because volume strength is the volume of oxygen at STP given by one volume sample of hydrogen peroxide on heating.

Question 1

The volume strength of  3N\:H_{2}O_{2}  solution is:

Accepted Answer

3\:N  means  3  g eqwt. in  1  L solution.  N=\dfrac{	ext{weight}}{	ext{eq. wt.}} . ( eq. wt. for hydrogen peroxide  = 17  g )  Therefore,  3 = \dfrac{	ext{wt}}{17}  or  	ext{wt} = 3 	imes17 = 51  g.Now, from the equation, we know that  68  g of  H_2O_2  gives  22.4  L of Oxygen. Therefore,  51  g of hydrogen peroxide gives  22.4 	imes \dfrac{51}{68} = 16.8  L.Thus, volume strength is  16.8  L, because volume strength is the volume of oxygen at STP given by one volume sample of hydrogen peroxide on heating.

Question 2

On Fahrenheit scale, the boiling point of water is:

Accepted Answer

On the Fahrenheit scale, the freezing point of water is  32^{\circ}F  and the boiling point is  212^{\circ}F  (at standard atmospheric pressure).Fahrenheit to Celsius :   (F - 32) * 1.8 =C  Boiling point of water is 100 degree C. Then,  F = (100 * 1.8) + 32  =  212^{\circ}F Option C is correct.

Question 3

Hydrogen resembles halogen in many respects for which several factors are responsible. Of the following factors which one is the most important in this respect?

Accepted Answer

Halogens like hydrogen are highly electronegative in nature as it has the tendency to form anion by the gain of one electron in its valence shell. Similarly hydrogen has one electron in its only s-orbital which can accommodate only one electron to form hydride anion like halogens.

Question 4

In what respect is the electronic configuration of hydrogen and halogens similar?

Accepted Answer

Hydrogen has one electron in its only shell. Hence, it needs one more electron to attain a stable duplet electronic configuration. On the other hand, halogens have seven electrons in their valence shell. They also need one more electron to attain a stable octet configuration. Therefore, they are similar in the way that they have one electron less than noble gas configuration. H-1s^1; He -1s^2 X-ns^2 np^5 : Ne-1s^22s^2 2p^6 Both have an electron less than the nearest noble gas configuration.