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GarryVolchara [31]
3 years ago
6

If you dilute 19.0 mL of the stock solution to a final volume of 0.310 L , what will be the concentration of the diluted solutio

n?
Chemistry
1 answer:
Dahasolnce [82]3 years ago
6 0

Answer:

M_2=0.613M_1

Explanation:

M_1 = Concentration of stock solution

M_2 = Concentration of solution

V_1 = Volume of stock solution = 19 mL

V_2 = Volume of solution = 0.31 L= 310 mL

We have the relation

M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{M_119}{310}\\\Rightarrow M_2=M_1\times\dfrac{19}{310}\\\Rightarrow M_2=0.613M_1

\boldsymbol{\therefore M_2=0.613M_1}

The concentration of the diluted solution will be 0.613 times the concentration of the stock solution.

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One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo
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Answer:

The total work is 4957.45J

Explanation:

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W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J

Stape 2: n  = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J

The total work is the sum of the two steps

W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J

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