# Question #a0387

##### 1 Answer

#### Answer:

The answer is **(2)**

#### Explanation:

The idea here is that you're going to observe a **smaller** molar mass for silver nitrate because some of the salt will *dissociate* when dissolved in water to produce silver(I) cations and nitrate anions in aqueous solution.

#"AgNO"_ (3(aq)) -> "Ag"_ ((aq))^(+) + "NO"_ (3(aq))^(-)#

Notice that **every mole** of silver nitrate that dissociates in solution produces **mole** of silver(I) cations and **mole** of nitrate anions, hence **moles** of ions. Keep this in mind.

Now, let's say that your sample has a mass of **moles** of silver nitrate.

You can express the **molar mass** of the salt, which represents the mass of exactly **mole** of silver nitrate, in terms of the mass

#M_M = m/n#

You calculate that silver nitrate has a molar mass of

#170 color(red)(cancel(color(black)("g mol"^(-1)))) = (m color(red)(cancel(color(black)("g"))))/(n color(red)(cancel(color(black)("moles"))))#

#170 = m/n" "color(orange)("(*)")#

At this point, you dissolve the sample in water. Let's say that *that dissociate* to produce ions.

You can say that the solution will contain

#1 - f * n -># the number of moles of silver nitrate thatdo not dissociate

#color(red)(2) * f * n -># the number of moles ofionsproduced in solution

The **total number of moles** of solute, dissociate *and* undissociated, present in the solution is equal to

#1 - fn + color(red)(2) * fn = 1 + fn#

At this point, you observe that the molar mass comes out to be

#92.64 color(red)(cancel(color(black)("g mol"^(-1)))) = (m color(red)(cancel(color(black)("g"))))/( (1 + fn)color(red)(cancel(color(black)("moles"))))#

#92.64 = m/(1 + fn)" "color(darkorange)("(* *)")#

Notice that we have two equations with *three unknowns*, the mass of the sample, the number of moles it contains, and the fraction that dissociates.

In order to be able to calculate the value of

To make the calculations easier, let's say that we're working with **molar mass** of the salt, you can say that

#"170 g AgNO"_3 " " -> " " "1 mole AgNO"_3#

At this point, you will have

#{(m= "170 g"), (n = "1 mole") :}#

Plug this into equation

#92.64 = 170/(1 + f * 1)#

#1 + f = 170/92.64#

#f = 170/92.64 - 1 = 0.835#

So, you know that a fraction equal to *dissociates*. To convert this to a **percentage**, multiply the value of

#color(darkgreen)(ul(color(black)("% dissociation" = 83.5%)))#