Answer:
Mole fraction for solute = 0.1, or 10%
Molality = 6.24 mol/kg
Explanation:
22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH
Mass of solution = 100 g
Mass of solute = 22.3 g
Mass of solvent = 100 g - 22.3g = 77.7 g
Let's convert the mass to moles
22.3 g . 1mol/ 46 g = 0.485 moles
77.7 g. 1mol / 18 g = 4.32 moles
Total moles = 4.32 moles + 0.485 moles = 4.805 moles
Xm for solute = 0.485 / 4.805 = 0.100 → 10%
Molality → mol/ kg → we convert the mass of solvent to kg
77.7 g. 1 kg / 1000g = 0.0777 kg
0.485 mol / 0.0777 kg = 6.24 m
Answer:
16 mol NaCl.
Explanation:
Do the train track method to cancel out all the units except moles of NaCl on top. Remember one mole of any gas occupies 22.4 L at STP.
179.2 L CO2 x 1 mol CO2/22.4 L CO2 x 2 mol NaCl/1 mol CO2
= 16 mol NaCl
