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jeka57 [31]
3 years ago
6

How many grams of sodium chloride must dissolve in 750.0 g of water to make a 0.50 molal solution?

Chemistry
1 answer:
AleksandrR [38]3 years ago
7 0
Molality is defined as the number of moles of solute in 1 kg of solvent.
molality of solution to be prepared is 0.50 molal
this means that in 1000 g of water there should be 0.50 mol of NaCl
if 1000 g of water should contain - 0.50 mol 
then 750.0 g of water requires - 0.50 mol/kg x 0.750 kg = 0.375 mol
mass of NaCl in 0.375 mol - 58.5 g/mol x 0.375 mol = 21.9 g
therefore a mass of 21.9 g of NaCl is required 
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Potassium-40 is a radioactive isotope that decays into a single argon-40 atom and other particles with a half-life of 1:25 billi
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Answer:

0.147 billion years = 147.35 million years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
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  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
  • The half-life of Potassium-40 is 1.25 billion years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.25 billion years) = 0.8 billion year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

<em></em>

where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).

[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).

  • At the time needed to be determined:

<em>8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.</em>

  • If we start with 100% Potassium-40:

∴ The remaining concentration of Potassium-40 ([A] = 88.88%).

and that of argon-40 produced from potassium-40 decayed = 11.11%.

  • That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).

∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.

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