Answer:
d) 0.1202 M
Explanation:
Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.
NaOH + HA → NaA + H₂O
The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:
22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol
The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.
The molar concentration of HA is:
2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M
Molality is defined as the number of moles of solute in 1 kg of solvent.
since density and volume has been given we can calculate the mass of water
mass = density x volume
mass = 1.00 g/mL x 1620 mL = 1620 g
number of moles of glucose - 257 g / 180 g/mol = 1.43 mol
number of moles of glucose in 1620 g - 1.43 mol
therefore number of moles in 1000 g - 1.43 mol / 1.620 kg = 0.883 mol/kg
therefore molality is 0.883 mol/kg
Answer:
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The correct answer of the given question above would be option D. MAGNESIUM PHOSPHITE. The name of the binary compound Mg3 (PO3)2 is magnesium phosphite. Magnesium phosphate is <span>Mg3(PO4)2. Magnesium Phosphide is </span><span>Mg3P2. Hope this answer helps. Thanks for posting your question. </span>
