Monosaccharides are the simplest carbohydrates. Although glucose and fructose have the same molecular formula they have different structures or the atoms are arranged differently from each other and this is evident in the way they react, behave and in their properties.
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
Answer:
Explanation:
the molecular mass of Na2C6H6O7 =236 g\mole it has a sodium that has 23 g/mole so 7.6 g of Na2C6H6O7 has x of sodium mass
236 g/mole ⇒ 23g/mole
<h2> 7.6 g ⇒ ˣ </h2>
7.6 x 23 ÷ 236 = 74.07×10-2 grams of sodium
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