Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
ΔT(freezing point) = (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.50 mol/kg)
ΔT(freezing point) = 0.93 °C
Tf - T = 0.93 °C
<span>T = -0.93 °C</span></span>
Answer: 27.27%
Mass percentage of carbon in carbon dioxide is 4412×100=27. 27%.
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Hopefully you get the correct answer I just need points
Ca=40
C=12
O=16
1 mole of CaCO3 has 100 grams
So 50 grams is 0.5 mole
