A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and of sulfate ions in the final solution.

NH

+

4

:

0.272

M

SO

2

−

4

:

0.136

M

Explanation:

We can solve this problem using some molarity calculations:

molarity

=

mol solute

L soln

We should convert the given mass of

(NH

4

)

2

SO

4

to moles using its molar mass (calculated to be

132.14

g/mol

):

10.8

g (NH

4

)

2

SO

4

⎛

⎝

1

l

mol (NH

4

)

2

SO

4

132.14

g (NH

4

)

2

SO

4

⎞

⎠

=

0.0817

mol (NH

4

)

2

SO

4

This is the quantity present in

100

mL soln

, so let's calculate the molarity of the solution (converting volume to liters):

molarity

=

0.0817

l

mol (NH

4

)

2

SO

4

0.100

l

L soln

=

0.817

M

10

mL

of this solution is added to

50

mL H

2

O

, which makes a

60

-

mL

total solution.

We can now use the dilution equation

M

1

V

1

=

M

2

V

2

to find the molality of the new,

60

-

mL

solution:

(

0.0817

M

)

(

10

l

mL

)

=

(

M

2

)

(

60

l

mL

)

M

2

=

(

0.817

M

)

(

10

mL

)

60

mL

=

0.136

M

This means that there are

0.136

moles of

(NH

4

)

2

SO

4

per liter of solution.

Let's recognize that

1

mol (NH

4

)

2

SO

4

contains

2

mol NH

+

4

1

mol SO

2

−

4

The concentrations of each ion is thus

(

2

)

(

0.136

M

)

=

0.272

M

NH

+

4

(

1

)

(

0.136

M

)

=

0.136

M

SO

2

−

4

Explanation:

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

if you're asking what comes out, it's steam.