Answer:
136.63 °C
Explanation:
ΔTb=Tb solution - Tb pure
Where; Tb pure = 133.60°C
molar mass of solute = 121.14 g/mol
number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles
molality = 0.431 moles/350 * 10^-3 = 1.23 molal
Then;
ΔTb = Kb * m * i
Kb = 2.46°C kg mol^-1
m = 1.23 molal
i = 1
ΔTb = 2.46 * 1.23 * 1
ΔTb = 3.03 °C
Hence;
Tb solution = ΔTb + Tb pure
Tb solution = 3.03 °C + 133.60°C
Tb solution = 136.63 °C
Answer:
63.55
Explanation:
relative atomic mass=(mass of isotope1×relative abundance)+(mass of isotope 2×relative abundance)/100
r.a.m=(62.93×69.09)+(64.93×30.91)/100
=(4347.8337)+(2006.9863)/100
=6354.82/100
=63.55
Answer: 18.0152 milliliters
Explanation:
Hi, to answer this question we have to apply the next formula:
Water volume = water mass / water density
Since 1 mol of water weights 18.0152 grams
Replacing with the values and solving:
Water volume = 18.0152 g / 1 g /ml = 18.0152 milliliters
Feel free to ask for more if needed or if you did not understand something.
