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
Oxidation of D -Ribose in presence of hypobromous acid gives D-Ribonic acid
P1V1/T1=P2V2/T2 P2=1/2 P1 V1=1 T1=298K
1 P1/298= (1/2) P1V2/373 cross P1
1/298=1/2V2/373
1/298=1/V2 746
v2=746/298
V2=2.5L
If we have 321 grams of a liquid, and the density is 0.84 g/mL, then we can easily find the volume of the liquid. We just need to take this 0.84 and multiply that by the number of grams. If we do 321 * 0.84, we get 269.64 mL. This is the volume that this liquid has.Remember this equation for future problems: V = D*M. V meaning volume, D meaning density, and M meaning mass. I hope this helps.
