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Arada [10]
2 years ago
13

When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2

.7 C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH CI) are dissolved in the same mass of X, the freezing point of the solution is 9.9 °C lower than the freezing point of pure X.
Required:
Calculate the van't Hoff factor for ammonium chloride in X.
Chemistry
1 answer:
Arlecino [84]2 years ago
5 0

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  =\dfrac{70.4 \ g}{121.14 \ g/mol}

= 0.58 mole

The molality = \dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent  }

= \dfrac{0.58 }{0.85 }

= 0.6837

Using the formula:

\mathbf {dT  = l   \times  k_f  \times m}

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = \dfrac{70.4 \ g}{53.491 \  g /mol}

= 1.316 mol

The molality = \dfrac{1.316 \ mol}{0.85 \ kg}

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = \dfrac{9.9}{3.949 \times 1.5484 \ m}

l = 1.62

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Answer:

A chemical equation can be described as a symbolic representation of a chemical reaction. It can be written in the form of words or symbols.

The following chemical reaction can be completed as :

Na2SO4·10H2O + 2C = Na2S + 10H2O + 2CO2

Sodium Sulfate Decahydrate + Diamond = Sodium Sulfide + Water + Carbon Dioxide

The products of the reaction are:

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6 0
3 years ago
Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =
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<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_{eq} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1 in the expression.

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2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)

The expression of K_c for above equation is:

K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}

We are given:

[H_2S]_{eq}=0.671M

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K_c=1.35

Putting values in above expression, we get:

1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}

[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M

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8 0
3 years ago
How many degrees of unsaturation are lost or gained in the base-induced transformation from mito-ph to its photoactive ring-open
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Unsaturation (IHD) 2 hydrogen Needed

IHD = [(2n+2) -H]/2
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Unsaturation:
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6 0
3 years ago
The heavier a gas molecule,
mr_godi [17]

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3 0
3 years ago
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-BARSIC- [3]

Answer:

132g/mole

Explanation:

using the formula PV=nRT should be used to solve for the number of moles (n).  R is a constant which is 62.3637 L mmHG/mole K.

Inorder for your units to match you will have to convert 125ml to .125L and the temperature of 85C to K . you do that by adding 273 to the 85C and get 358K.  Once you solve for n then you use that number and divide by the number of grams from the question (.560g) since molar mass is grams/moles.

4 0
3 years ago
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