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stealth61 [152]
3 years ago
10

Calculate molality,molarity and mole fraction of KI if the density of 20%(mass) aqueous KI is 1.202 g/mL

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
5 0
Basis: 1 L of the substance.
               (1.202 g/mL) x (1000 mL) = 1202 g 
                   mass solute = (1202 g) x 0.2 = 240.2 g
                   mass solvent = 1202 g x 0.8 =  961.6 g
                   moles KI = (240.2 g) x (1 mole / 166 g)  = 1.45 moles
                   moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent 
                  = 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
                  = 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
                        =  1.45 moles / (1.45 moles + 53.42 moles) = 0.0264 

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Explanation:

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2 years ago
The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.
Brilliant_brown [7]

Answer :  The rate of reaction is,

Rate=4.77\times 10^{-19}M/s

The appearance of NO_3 is, 4.77\times 10^{-19}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.

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NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)

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So, \text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s

Thus, the appearance of NO_3 is, 4.77\times 10^{-19}M/s

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