25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce 
g of (NH₄)₂S
25g of NH₃ will produce 
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S
The density of the sample is:
Density = mass / volume
Density = 9.85 / 0.675
Density = 14.6 g/cm³
If the sample has 95% gold, and 5% silver, its density should be:
0.95 x 19.3 + 0.05 x 10.5
Theoretical density = 18.9 g/cm³
The difference in theoretical and actual densities is very large, making it likely that the jeweler was not telling the truth.
We will use the formula for freezing point depression :
but first, we need to get the molality m of the solution:
- molality m = moles of C2H5OH / mass of water Kg
when moles of C2H5OH = mass of C2H5OH/ molar mass of C2H5OH
= 11.85 g / 46 g/mol
= 0.258 moles
and when we have the mass of water Kg = 0.368 Kg
so, by substitution on the molality formula:
∴ molality m = 0.258 moles / 0.368 Kg
= 0.7 mol/Kg
and when C2H5OH is a weak acid so, there is no dissociation ∴ i = 1
and when Kf is given = 1.86 C/m
so by substitution on ΔTf formula:
when ΔTf = i Kf m
∴ ΔTf = 1 * 1.86C/m * 0.7mol/Kg
= 1.302 °C
