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miskamm [114]
3 years ago
14

What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
4 0
We will use the expression for freezing point depression ∆Tf
     ∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is 
     ∆Tf = 0C - (-3°C) = 3°C 
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and  one Cl- ion:
     KCl → K+ + Cl- 
Therefore, the molality m of the solution can be calculated as 
     3 = 2 * 1.86 * m
     m = 3 / (2 * 1.86)
     m = 0.80 molal
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Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
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Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

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