How many `Na^+` ions are present in 100 mL of 0.25 M of NaCl solution ?

Options

A. `0.025xx 10^(23)`
B. `1. 505 xx 10^(22)`
C. ` 15 xx 10^(22)`
D. `2.5 xx 10^(23)`

Answer

B

Explanation

No.of moles of NaCl` = ( M xx V )/( 1000)`
`=(0.25 xx 100) /(1000) =0.025 `
`NaCl rarr Na^(+) +Cl^(-)`
No. of moles of `Na^(+)` ions = 0.025
No. of `Na^(+)` ions `=0.025 xx 6.023 xx10^(23) =1.505 xx 10^(22)`