Question

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

• A. $0.25\times {10}^{23}$
• B. $1.505\times {10}^{22}$
• C. $15\times {10}^{22}$
• D. $2.5\times {10}^{23}$

Answer 1

Answer: (B)

$1.505\times {10}^{22}$

No of moles (n) = $\frac{Molarity\times V_{mL}}{1000}$

In this problem

$V_{mL}=100mL$

$M=0.25M$

therefore

$n=\frac{0.25\times 100}{1000}$

=$\frac{25}{1000}=0.025mol$

therefore no of $N{a}^{+}$ ions = $n\times 6.022\times {10}^{23}$

= $0.025\times 6.022\times {10}^{23}$

= $1.505\times {10}^{22}ions$.