0.250 L*3M=0.250 L*3mol/L= 0.750 mol
To convert the given value to the desired one, use the proper unit conversions and dimensional analysis. Use the following conversion for the first set.
1 g = 100 cg
1 L = 1000 mL
Using the concept presented above,
V = (59800 cg/L)(1 g/100 cg)1 L/1000 mL)
V = 0.598 g/mL
Answer:
0.200 m K3PO3
Explanation:
Let us remember that the freezing point depression is obtained from the formula;
ΔTf = Kf m i
Where;
Kf = freezing point constant
m = molality
i = Van't Hoff factor
The Van't Hoff factor has to do with the number of particles in solution. Let us consider the Van't Hoff factor for each specie.
0.200 m HOCH2CH2OH - 1
0.200 m Ba(NO3)2 - 3
0.200 m K3PO3 - 4
0.200 m Ca(CIO4)2 - 3
Hence, 0.200 m K3PO3 has the greatest van't Hoff factor and consequently the greatest freezing point depression.
First, find out how many grams are in one mole of CO2(the two oxygen atoms means you need to multiply oxygen’s amu by 2,then add whatever carbon’s amu is to that). Then divide 26 grams by that number and that will be your moles. There are only two significant figures, so round your answer correctly.
