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lesya692 [45]
3 years ago
10

A solution is prepared by mixing 2.17 g of an unknown non-electrolyte with 225.0 g of chloroform. The freezing point of the resu

lting solution is –64.2oC. The freezing point of pure chloroform is – 63.5oC and its kf = 4.68oC m-1. What is the molecular mass of the unknown?
Chemistry
1 answer:
Deffense [45]3 years ago
6 0

Answer:

The molar mass of the unknown non-electrolyte is 64.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown non-electrolyte = 2.17 grams

Mass of chloroform = 225.0 grams

The freezing point of the resulting solution is –64.2 °C

The freezing point of pure chloroform is – 63.5°C

kf = 4.68°C/m

Step 2: Calculate molality

ΔT = i*kf*m

⇒ ΔT = The freezing point depression = T (pure solvent) − T(solution) = -63.5°C + 64.2 °C = 0.7 °C

⇒i = the van't Hoff factor = non-electrolyte = 1

⇒ kf = the freezing point depression constant = 4.68 °C/m

⇒ m = molality = moles unknown non-electrolyte / mass chloroform

0.7 °C = 1 * 4.68 °C/m * m

m = 0.150 molal

Step 3: Calculate moles unknown non-electrolyte

molality = moles unknown non-electrolyte / mass chloroform

Moles unknown non-electrolyte = 0.150 molal * 0.225 kg

Moles unknown non-electrolyte = 0.03375 moles

Step 4: Calculate molecular mass unknown non-electrolyte

Molar mass = mass / moles

Molar mass = 2.17 grams / 0.03375 moles

Molar mass = 64.3 g/mol

The molar mass of the unknown non-electrolyte is 64.3 g/mol

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1. If we have a reaction of Zn3PO4 and HCl, what would the products of this
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Answer:

1. The products of this reaction are ZnCl₂ and H₃PO₄.

2. 14.57 g.

Explanation:

<em>1. What would the products of this  reaction be?</em>

  • The balanced reaction between Zn₃(PO₄)₂ and HCl is represented as:

<em>Zn₃(PO₄)₂ + 6HCl → 3ZnCl₂ + 2H₃PO₄,</em>

It is clear that 1.0 mol of Zn₃(PO₄)₂ reacts with 6.0 mol of HCl to produce 3.0 mol of ZnCl₂ and 2.0 mol of H₃PO₄.

So, the products of this reaction are ZnCl₂ and H₃PO₄.

<em>2. If we produced 13.05 g of H₃PO₄, how many grams of  hydrochloric acid would be need to start with?​</em>

  • Firstly, we should get the no. of moles (n) of 13.05 grams of H₃PO₄:

n = mass/molar mass = (13.05 g)/(97.994 g/mol) = 0.1332 mol.

<u><em>Using cross-multiplication:</em></u>

6.0 mol of HCl needed to produce → 2.0 mol of H₃PO₄, from stichiometry.

??? mol of HCl needed to produce → 0.1332 mol of H₃PO₄.

∴ The no. of moles of HCl needed = (6.0 mol)(0.1332 mol)/(2.0 mol) = 0.3995 mol.

∴ The mass of HCl needed = n*molar mass = (0.3995 mol)(36.46 g/mol) = 14.57 g.

<em>So, the grams of  hydrochloric acid would be need to start with = 14.57 g.</em>

3 0
3 years ago
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Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

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The concentration of Pb(ClO₃)₂ is:

\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M

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