Respuesta :
My teacher just explained this problem to me and she said it is "the same".
Answer: the same
Explanation:
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock solution = 0.50 M
[tex]V_1[/tex] = volume of stock solution = 10 ml= 0.01 L (1L=1000ml)
[tex]M_2[/tex] = molarity of resulting solution = ? M
[tex]V_2[/tex] = volume of resulting solution = 100 ml = 0.1 L
[tex](0.50M)\times 0.01=(M_2)\times (0.1ml)[/tex]
[tex]M_2=0.05M[/tex]
Therefore, the Molarity of resulting solution is 0.05 M
[tex]Molarity=\frac{moles}{\text {Volume in L}}[/tex]
1. when volume is 10 ml or 0.01L (1L=1000ml)
moles of [tex]NaCl=Molarity\times {\text {Volume in L}}=0.50\times 0.01=5\times 10^{-3}moles[/tex]
2. when volume is 100 ml or 0.1L (1L=1000ml)
moles of [tex]NaCl=Molarity\times {\text {Volume in L}}=0.05\times 0.1=5\times 10^{-3}moles[/tex]
Thus they contain same amount of [tex]NaCl[/tex].
