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leonid [27]
3 years ago
14

What is the molality of a solution containing 125 grams of iodine (I2) and 750 grams of CCl4? A. 1.2 m. . B. 6.57 m. . C. 0.657

m. . D. 0.12 m.
Chemistry
1 answer:
uysha [10]3 years ago
4 0
The answer is <span>C. 0.657 m.
</span>
The molality is the number of moles of solute divided by kilograms of solvent. In a solution, solute the component which is present in lesser amount while solvent is present in greater amount. Thus, I2 is solute, and CCl4 is solvent.

So, first we need to calculate moles of I2:
Atomic mass of iodine: A(I) = 126.9
Molecular mass of iodine: Mr(I2) = 2*A(I) = 2 * 126.9 = 253.8
This means there are 253.8 g of iodine in 1 mole. So, 125 g of iodine will be in how many moles?
253.8 : 1 = 125 : x         ⇒      x = 125/253.8 = 0.493 moles

Now, let's convert amount of CCl4 in kilograms (1 kg = 1000 g):
750 g = 750/1000 kg = 0.75 kg

And now, it's only left to calculate molality of the solution:
numbers of moles of solute/kilograms of solvent = 0.493 moles/0.75 kg = 0.657 m
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Consider the reaction 5 Br− (aq) + BrO3− (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l) The average rate of consumption of Br− is 1.8
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Answer :  The average rate of consumption of H^+ during the same time interval is, 2.17 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

As we are given:

\frac{d[Br^-]}{dt}=1.81M/s

Now we have to determine the average rate of consumption of H^+ during the same time interval.

As,

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

or,

-\frac{1}{6}\frac{d[H^+]}{dt}=-\frac{1}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times (1.81M/s)

\frac{d[H^+]}{dt}=2.17M/s

Thus, the average rate of consumption of H^+ during the same time interval is, 2.17 M/s

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