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Alik [6]
2 years ago
14

A beaker contains 0.125 L of a 3.00 M solution. If the volume goes up to 0.325 L, what is the new molarity?

Chemistry
1 answer:
igor_vitrenko [27]2 years ago
8 0

Answer:

1.15 M

Explanation:

Step 1: Given data

  • Initial volume (V₁): 0.125 L
  • Initial concentration (C₁): 3.00 M
  • Final volume (V₂): 0.325 L
  • Final concentration (C₂): ?

Step 2: Calculate the final concentration of the solution

We want to prepare a dilute solution from a concentrated one by adding water. We can calculate the concentration of the dilute solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 3.00 M × 0.125 L/0.325 L = 1.15 M

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Answer:

9.15 atm

Explanation:

Ideal gas equation of state PV=nRT

P in hPa, V in L, n in mol, R is a constant which is 83.1 hpa*L/mol*k, T in kelvin.

Plug in all the number, and we will get:

P*6.21=2.02*83.1*343

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An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for
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Answer: The freezing point and boiling point of the solution are -6.6^0C and 101.8^0C respectively.

Explanation:

Depression in freezing point:

T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_f = freezing point of solution = ?

T^o_f = freezing point of water = 0^0C

k_f = freezing point constant of water = 1.86^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_f=-6.6^0C

Therefore,the freezing point of the solution is -6.6^0C

Elevation in boiling point :

T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of water = 100^0C

k_b = boiling point constant of water = 0.52^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_b=101.8^0C

Thus the boiling point of the solution is 101.8^0C

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Here we will use the general formula of Nernst equation:

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and R is a constant = 8.314 J/mol K

and T is the temperature in Kelvin = 73 + 273 = 346 K

and F is Faraday's constant = 96485 C/mole

and n is the number of moles of electron transferred in the reaction=2  

and Q is the reaction quotient for the reaction 
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