Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point
elevation is directly proportional to the molality of the solution
according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point
elevation.
Kb - the ebullioscopic
constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
The molecular weight for Calcium Chloride Dihydrate (CaCl2•2H20) is 2H2O.
mass of 64g/mol = 256.52 g/mol
mass of 11g/mol = 26.981538 g/mol
mass of 147g/mol= 146.9149 g/mol
258g/mol = 258.09843 g/mol
The correct answer is base
Answer:
28.01g
Explanation:
Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.
The molecular weight of a compound can be gotten by adding the molar weights of the elements that constitutes the compound .
The molecular weight of the compound CO is therefore
equal to the sum of the weight of both elements.
That’s = 12.01g + 16.00g
= 28.01g
Therefore, the molecular weight of CO is 28.01g
