Me i dont have much will give ten points in an experiment, two unknown compounds (one an alcohol and the other an ether) of equal molecular mass were dissolved in water. the result of the experiment is shown in the table. solubility comparison unknown compound solubility (g/100 ml water) unknown compound a--> solubility is 7.7 unknown compound b--> solubility is 6.8 which of the following correctly explains the identity of compound b and its solubility? a. it is an alcohol because the lack of hydrogen bonding between the oh^- make is less soluble b. it is an ether because it is unable to to form a hydrogen bond, so it is less soluble than water c. it is an ether because in the absence of a highly electronegative atom like o, f, or n, it lacks hydrogen bonding and thus makes it less soluble d. it is an alcohol because dispersion forces between the oh^- group and organic chain make it less soluble.

For the compound B the following statement is correct-

B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.

The solubility of alcohol in water depends upon the capability of formation of hydrogen bond in the solute. Now in alcohol the -OH group is polar in nature which enhance the possibility of hydrogen bond formation and it is more soluble in water.

On the other hand although there presence a -O- functional group in ether. It is less soluble in water due to non polarity of the functional group.

From the given data it is seen that compound A is more soluble in water than compound B. Thus it may be predicted that compound A is alcohol and B is ether.

Henceforth, for the compound B the following statement is correct-

B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.

The reason of incorrect options:

A. compound B cannot be an alcohol as it is less soluble in water.

C. In ether the functional group is -O-, thus electronegative atom (O) is present.

D. As both the compound (alcohol and ether) has equal molecular mass thus the organic chain will be same in alcohol and the hydrogen bond interaction will be more prominent than the dispersion force between the -OH group.  

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g