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gizmo_the_mogwai [7]
3 years ago
13

The lab just ran out of 1 M HCl that you need to complete the Benzillic Acid lab. The TA tells you there is 12 M HCl in the fume

hood. You need to make 10.0 mL of 1.0 M HCl. What volume or 12 M HCl will you need to dilute to get the correct volume and concentration of 1.0 M HCl?
Chemistry
1 answer:
Viktor [21]3 years ago
7 0

Answer:

0.83 mL

Explanation:

Given data

  • Initial concentration (C₁): 12 M
  • Initial volume (V₁): ?
  • Final concentration (C₂): 1.0 M
  • Final volume (V₂): 10.0 mL

We can calculate the initial volume of HCl using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 1.0 M × 10.0 mL / 12 M

V₁ = 0.83 mL

The required volume of the initial solution is 0.83 mL.

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Pete Gannett

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Ph.D. Chemistry, University of Wisconsin-Madison, (1982)2y

Seems to be an ideal gas law question. The relevant equation is:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles of gas, R is the gas constant (0.082 atm-L/mole-deg K), and T is temperature in Kelvins. STP means standard temperature and pressure and this is taken as 1 atm and 0º C or 273 K.

To calculate the number of molecules we will use the constant 6.023 * 10^23 molecules/mole and, therefore, we will need to know the number of moles (n). So, first we’ll rearrange the gas law equation, isolating ’n’ and then put the numbers in.

n = PV/RT = 1 * 1 / (0.082)(273) = 0.0447 moles

So, to calculate the number of molecules, multiple this by the number of molecules in a mole and you get:

# molecules of nitrogen in 1 Liter at STP = 6.023 * 10^23 molecules/mole * 0.0447 moles = 2.6905 * 10^22 molecules

Note, it does not matter what the gas is.

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