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Nuetrik [128]
3 years ago
14

The freezing point of 42.19 g of a pure solvent is measured to be 43.17 ºC. When 2.03 g of an unknown solute (assume the van 't

Hoff factor = 1.0000) is added to the solvent the freezing point is measured to be 40.32 ºC. Answer the following questions ( the freezing point depression constant of the pure solvent is 7.67 ºC·kg solvent/mol solute).What is the molality of the solution? m
How many moles of solute are present? mol
What is the molecular weight of the solute? g/molIf the freezing point depression constant of the pure solvent is 7.66 ºC·kg solvent/mol solute calculate the molar mass of the solute. g/mol
Chemistry
1 answer:
vovikov84 [41]3 years ago
4 0

Answer:

The molality of the solution is 0.3716 mol/kg

The number of moles of solute is 0.0157 mol

The molecular weight of the solute is 129.30 g/mol

The molar mass of the solute is 129.32 g/mol

Explanation:

m (molality of the solution) = ∆T/Kf = (43.17 - 40.32)/7.67 = 0.3716 mol/kg

Number of moles of solute = molality × mass of solvent in kilogram = 0.3716 × 0.04219 = 0.0157 mol

Molecular weight of solute = mass/number of moles = 2.03/0.0157 = 129.3 g/mol

When Kf = 7.66 °C.kg/mol

Molar mass = 2.03 ÷ (2.85/7.66 × 0.04219) = 129.32 g/mol

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3 years ago
1. For every 5.00 mL of milk of magnesia there are 400. mg of magnesium hydroxide. How many mL of milk of magnesia do we need to
forsale [732]

Answer:

1. 6.50mL

2. pH = 13.097

Explanation:

1. The neutralization of HCl with Al(OH)₃ is:

3HCl + Al(OH)₃ → Al(Cl)₃ + 3H₂O

40.0mL ≡ 0.0400L of 0.500M HCl are:

0.0400L × (0.500mol / L) = <em><u>0.0200 moles of HCl</u></em>

Based on the neutralization, 3 moles of HCl react with 1 mole of Al(OH)₃, thus, the moles of Al(OH)₃ that react with 0.0200 mol of HCl are:

0.0200mol HCl × (1mol Al(OH)₃ / 3mol HCl) = <em><u>0.00667 moles of Al(OH)₃</u></em>. In grams:

0.00667 moles Al(OH)₃ × (78g / 1mol) = 0.520g of Al(OH)₃ ≡ 520mg

As 5.00mL of milk magnesia contain 400mg of Al(OH)₃, the mL of milk magnesia required for a complete reaction are:

520 mg Al(OH)₃ × (5.00mL / 400mg) =<em> 6.50mL</em>

<em />

2. The reaction of lithium hydroxide (LiOH) with perchloric acid (HClO₄) is:

HClO₄ + LiOH → LiClO₄ + H₂O

<em>The reaction is 1:1.</em>

Moles of LiOH and HClO₄ are:

LiOH: 0.0500L × (0.350mol / L) = <em>0.0175 moles of LiOH</em>

HClO₄: 0.0300L × (0.250mol / L) = <em>0.0075 moles of HClO₄</em>

Assuming the reaction goes to completion, moles of LiOH that remains are:

0.0175 mol - 0.0075 mol = 0.0100 moles of LiOH. The total volume is 80.0mL, 0.0800L. Thus, molarity of LiOH is:

0.0100 mol / 0.0800L = <em>0.125 M of LiOH</em>

It is possible to obtain the pOH of the solution thus:

pOH = -log (OH) = -log 0.125M = 0.903

As pH = 14- pOH,

pH = 14 - 0.903 = <em>13.097</em>

I hope it helps!

3 0
2 years ago
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