Describe, in detail, how to make 5.5 liters of a 2.5 molar HCl solution from a 10.5 molar HCl stock solution.

1 answer:

5 0

The stock solution contains 10.5 moles of HCl per litre. A 5.5 litre solution of 2.5M HCl contains 5.5x2.5 = 13.75moles of HCl. Since every litre of stock solution provides 10.5M HCl, the amount of stock solution needed is 13.75/10.5 = 1.309L. Therefore you would dilute 1.309L of stock solution to 5.5L

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What’s the answer pls hrlp

arsen [322]

Answer:

The answer is A

Explanation:

5 0

3 years ago

A sample of neon occupies a volume of 752 ml at 25 0

lyudmila [28]

$V_{initial} = 752\:mL$
$T_{initial} = 25.0^0C$
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → $T_{initial} = 298.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 50.0^0C$
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → $T_{final} = 323.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }$
Product of extremes equals product of means:
$298.0* V_{f} = 752*323.0$
$298.0 V_{f} = 242896$
$V_{f} = \frac{242896}{298.0}$
$\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark$