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miv72 [106K]
3 years ago
8

A 74.28-g sample of ba(oh)2 is dissolved in enough water to make 2.450 liters of solution. how many ml of this solution must be

diluted with water in order to make 1.000 l of 0.100 m ba(oh)2?
Chemistry
1 answer:
Marianna [84]3 years ago
5 0

First let us calculate the initial molarity of the 2.45 L of solution. Molar mass = 171.34 g/mol

<span>moles Ba(OH)2 = 74.28 g * (1 mole / 171.34 g) =  0.4335 moles</span>

Molarity (M1) = 0.4335 moles / 2.45 L = 0.177 M

 

Now using the formula M1V1 = M2V2, we can calculate how much to dilute (V1):

0.177 * V1 = 0.1 * 1

V1 = 0.56 L

 

<span>Therefore 0.56 L of the initial solution must be diluted to 1 L to make 0.1 M</span>

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PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

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

-----> T = temperature (K)

Because density is comparing the mass per 1 liter, I am assuming that the system has a volume of 1 L. Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.

P = 1.00 atm                         R = 0.0821 L*atm/mol*K

V = 1.00 L                             T = 25.0. °C + 273.15 = 298.15 K

n = ? moles

PV = nRT

(1.00 atm)(1.00L) = n(0.0821 L*atm/mol*K)(298.15 K)

1.00 = n(0.0821 L*atm/mol*K)(298.15 K)

1.00 = (24.478115)n

0.0409 = n

Now, we need to find the molar mass using the number of moles per liter (calculated) and the density.

0.0409 moles           ? grams           4.95 grams
----------------------  x  ------------------  =   ------------------
        1 L                       1 mole                     1 L

? g/mol = 121 g/mol

**note: I am not 100% confident on this answer

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