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lesya692 [45]
3 years ago
5

Calculate the new boiling point of a solution if 10.00 g of a non-ionizing compound (C3H5(OH)3) is dissolved in 90.00 g of H2O.

Molar Mass of C3H5(OH)3 = 92.09 g/mol Kb = 0.51 oC/m (Answer must be in 4 sig fig. Do not include units in your answer).
Chemistry
1 answer:
erma4kov [3.2K]3 years ago
5 0

Answer:

Boiling T° of solution = 100.6

Explanation:

Formula for elevation of boiling point is:

ΔT = Kb . m . i

where ΔT means Boiling T° of solution - Boiling T° of pure solvent

Our solute is a non ionizing compound.

i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.

m = molality (moles of solute dissolved in 1 kg of solvent)

90 g of solvent = 0.09 kg of solvent

We convert mass of solute to moles (by the molar mass):

10 g . 1 mol /92.09 g = 0.108 moles

m = 0.108 mol /0.09 kg = 1.21 m

Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1

Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C

Boiling T° of solution = 100.6

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Briefly describe four ways that a beam can be strengthened?​
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1. Changing Beam Material

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3. Changing Beam form

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Consider the following reaction between mercury(II) chloride and oxalate ion.
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<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
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