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

Options

A. `0.025 xx 10^23`
B. `1.505xx10^22`
C. `15xx10^22`
D. `2.5xx10^23`

Answer

B

Explanation

No. of moles of NaCl =`(MxxV)/1000=(0.25xx100)/(1000)=0.025`
`NaCl to Na^(+) + Cl^(-)`
No. of moles of `Na^+` ions = 0.025
No. of `Na^+` ions = 0.025 x 6.023 x `10^23` = `1.505xx10^22`