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How many moles are in 1.93x1023 particles of Na3PO4?

User Pyjamas
by
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2 Answers

2 votes

Answer :

No. of particles of Na₃PO₄ = 1.93×10²³

We have to find number of moles of the given compound
.

Number of moles cam be found by using this formula :


\bigstar\:\boxed{\bf{\purple{n=(N)/(N_A)}}}

  • n denotes number of moles
  • N denotes number of particles

  • \sf{N_A}denotes avogadro's constant

We know that,


  • \bf{N_A} = 6.022×10²³

Calculation :

➠ n = N/
\sf{N_A}

➠ n = (1.93×10²³)/(6.022×10²³)

n = 0.32 moles

Explore More :

  • n = weight / Molar mass
  • n = volume / 22.4

Hope It Helps!

User Lewis Lebentz
by
7.5k points
1 vote

Answer:

The answer is 0.32 moles

Step-by-step explanation:

To find the number of moles given the number of entities we use the formula


n = (N)/(L) \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

N = 1.93 × 10²³ particles

We have


n = \frac{1.93 * {10}^(23) }{6.02 * {10}^(23 ) } \\ = 0.320598006...

We have the final answer as

0.32 moles

Hope this helps you

User Arkh
by
8.1k points

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