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Question: How many ![](https://cdn.pureessence.tech/canvas_254.png?top_left_x=1563&top_left_y=618&width=300&he...

How many ions are present in 100 mL of 0.25M0.25 M of NaCl solution

A

0.025×10230.025 \times 10 ^ { 23 }

B

1.505×10221.505 \times 10 ^ { 22 }

C

15×102215 \times 10 ^ { 22 }

D

2.5×10232.5 \times 10 ^ { 23 }

Answer

1.505×10221.505 \times 10 ^ { 22 }

Explanation

Solution

No. of moles of NaCl

=M×V1000=0.25×1001000=0.025= \frac { \mathrm { M } \times \mathrm { V } } { 1000 } = \frac { 0.25 \times 100 } { 1000 } = 0.025

NaCl→Na++Cl−\mathrm { NaCl } \rightarrow \mathrm { Na } ^ { + } + \mathrm { Cl } ^ { - }

No. of moles of Na+\mathrm { Na } ^ { + }ions =0.025= 0.025

No of Na+\mathrm { Na } ^ { + }ions = 0.025×6.023×1023=1.505×10220.025 \times 6.023 \times 10 ^ { 23 } = 1.505 \times 10 ^ { 22 }