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How many `Na^+` ions are present in 100 mL of 0.25 M of NaCl solution ?A. `0.025 xx 10^23`B. `1.505xx10^22`C. `15xx10^22`D. `2.5xx10^23`

Answer» Correct Answer - B
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`


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