Respuesta :
Answer: B. 0.015 mole/L
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of [tex]NaHCO_3[/tex] solution = 0.010 M
Volume of solution = 10 mL
Putting values in equation 1, we get:
a) [tex]0.010M=\frac{\text{Moles of}NaHCO_3\times 1000}{10ml}\\\\\text{Moles of }NaHCO_3=\frac{0.010mol/L\times 10}{1000}=10^{-4}mol[/tex]
1 mole of [tex]NaHCO_3[/tex] contains = 1 mol of [tex]Na^+[/tex]
Thus [tex]10^{-4}mol[/tex] of [tex]NaHCO_3[/tex] contain= [tex]\frac{1}{1}\times 10^{-4}=10^{-4}[/tex] mol of [tex]Na^+[/tex]
b) [tex]0.010M=\frac{\text{Moles of}Na_2CO_3\times 1000}{10ml}\\\\\text{Moles of }Na_2CO_3=\frac{0.010mol/L\times 10}{1000}=10^{-4}mol[/tex]
1 mole of [tex]Na_2CO_3[/tex] contains = 2 mol of [tex]Na^+[/tex]
Thus [tex]10^{-4}mol[/tex] of [tex]NaHCO_3[/tex] contain= [tex]\frac{2}{1}\times 10^{-4}=2\times 10^{-4}[/tex] mol of [tex]Na^+[/tex]
Total [tex][Na^+]=\frac {\text {total moles}}{\text {total volume}}=\frac{10^{-4}+2\times 10^{-4}}{0.02L}=0.015M[/tex]
The molar concentration of [tex]Na^+[/tex] in a solution is 0.015 M
