How many ammonium ions and how many sulfate ions are present in an 0.260 mol sample of (nh4)2so4?

1 answer:

4 0

The question requires that you assume complete ionization of the compound. This is:

(NH4)2 SO4 → 2 NH4 (+) + SO4 (2-)

Then, 1 mol of (NH4)2 SO4 produces 3 moles of ions: 2 moles of NH4 (+) and 1 mol of SO4 (2-)

=> 0.26 * 2 moles of NH4(+) = 0.52 moles of NH4 (+)

0.26 * 1 mol of SO4 (2-) = 0.26 moles of SO4 (2-).

Answer: 0.52 moles of ammonium ions and 0.26 moles of sulfate.ions.

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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$