To calculate the mass of sodium sulfate required, we need to follow these steps:
1. Convert the volume of silver nitrate solution to moles.
2. Calculate the moles of silver that need to be precipitated.
3. Calculate the moles of sodium sulfate required to precipitate this amount of silver.
4. Convert the moles of sodium sulfate to mass.
Let's start with step 1:
To convert from volume to moles, we need to use the formula V = nRT/P, where V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm mol⁻¹ K⁻¹), T is the temperature in Kelvin, and P is the pressure in atmospheres. In this case, the pressure and temperature can be taken as atmospheric pressure (1 atm) and room temperature (298 K), respectively. The 2.25 M concentration of silver nitrate means that each liter of solution contains 2.25 mol of silver nitrate. So the amount of silver nitrate in a 75 mL sample is :
75 mL * (2.25 mol/L) = 169.52 mol
Now let's proceed with step 2:
To calculate the moles of silver that need to be precipitated, we need to consider the reaction formula. Here's how it goes:
AgNO3 + Na2SO4 → Ag2SO4 + NaNO3
We need to find out how much Na2SO4 is required to react with all the AgNO3. To do this, we need to calculate the moles of Na2SO4 that react with 1 mole of AgNO3. This is given by the stoichiometric coefficient for Na2SO4, which is 1 in this case. Thus we need:
169.52 mol AgNO3 * (1 mol Na2SO4 / 1 mol AgNO3) = 169.52 mol Na2SO4
Finally, we need to calculate the mass of Na2SO4 required. We can do this by multiplying the molarity (2.25 M) by the volume (75 mL):
169.52 mol * 2.25 mol/L * 0.075 L = 31.2 g
Therefore, 3