Question

Number of ions present in 100 ml, 0.5 M NaCl solution are 100 ​

Answer 1

Explanation:

Number of moles of NaCl=

1000

MV

=

1000

0.5×50

=0.025

NaCl→Na

+

+Cl

−

Number of moles of Na

+

= Number of moles of NaCl

=0.025

Number of ions of Na

+

=0.025×6.023×10

23

=1.505×10

22

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Answer 2

Answer:

No.of sodium or chloride ions are

$3.011 \times 10 {}^{22}$

Explanation:

Given that,

Molarity of sodium chloride solution is 0.5M

The volume of the solution is 100ml=0.1L

To find number of ions,we need to find no.of moles which can be found by below relation

$molarity = \frac{moles}{volume(in \: lit)}$

Substituting known values,we get

$0.5 = \frac{n}{0.1} = > n = 0.05 \: mol$

No.of moles are known.Therefore no.of sodium or chloride ions are found as

$0.05 \times 6.023 \times 10 {}^{23} \\ = 3.011 \times 10 {}^{22}$

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