Ionic molecules have higher boiling point than covalent molecules.
So that leaves Na2SO4 and NaCl.
The ionic molecule with higher charge will have higher boiling point.
Na2SO4 have ions with charge +-2.
NaCl have ions with charge +-1.
So the answer is C.
To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
= 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
mass of NH3 = no.moles * molar mass of ammonia
= 14.28 * 17 = 242.76 g
This problem is simply converting the concentration from molality to molarity. Molality has units of mol solute/kg solvent, while molarity has units of mol solute/L solution.
2.24 mol H2SO4/kg H2O * (0.25806 kg H2SO4/mol H2SO4) = 0.578 kg H2SO4/kg H2O
That means the solution weighs a total of 1 kg + 0.578 kg = 1.578 kg. Then, convert it to liters using the density data:
1.578 kg * (1000g / 1kg) * (1 mL/1.135 g) = 1390 mL or 1.39 L.
Hence, the molarity is
2.24/1.39 = 1.61 M