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Marianna [84]
2 years ago
12

1. How many moles of NaCl would be contained in .750 L solution with a molarity of 0.45M?

Chemistry
2 answers:
Evgesh-ka [11]2 years ago
5 0
Formula of Molarity =

Molarity = no. of moles /volume of solution in L

Given, 0.45M= no. of moles/0.750L
Therefore, no. of moles = 0.45M x 0.750L= 0.3375
yarga [219]2 years ago
4 0

Answer:

\boxed {\boxed {\sf 0.3375 \ mol \ NaCl}}

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

M=\frac{moles \ of \ solute }{liters \ of \ solution}

We know the molarity is 0.45 M and there are 0.750 liters of solution. The solute is NaCl (sodium chloride). We can substitute the values into the formula.

0.45 \ M = \frac{ moles \ of \ NaCl}{ 0.750 \ L }

The molarity (M) can also be represented by mol/L

0.45 \ mol/L = \frac{ moles \ of \ NaCl}{ 0.750 \ L }

We are solving for the moles of solute, so we must isolate the numerator. It is being divided by 0.750 liters. The inverse operation is multiplication, so multiply both sides of the equation by 0.750 L.

0.750 \ L * 0.45 \ mol/L = \frac{ moles \ of \ NaCl}{ 0.750 \ L }*0.750 \ L

The liters will cancel out.

0.750  * 0.45 \ mol = { moles \ of \ NaCl}

0.3375 \ mol \ = moles \ of NaCl

There are <u>0.3375 moles of NaCl</u> in a 0.750 liter solution with a molarity of 0.45 M.

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a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

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a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

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ΔHºrxn = 116.3 kJ

Similarly,

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b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

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0 = 116 kJ -T (0.113 kJ/K)

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ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

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