Answer:
"0.60 g" is the appropriate solution.
Explanation:
The given values are:
Volume of base,
= 30 ml
Molarity of base,
= 0.05 m
Molar mass of acid,
= 400 g/mol
As we know,
⇒ [tex]Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}[/tex]
On substituting the values, we get
⇒ [tex]0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}[/tex]
⇒ [tex]Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}[/tex]
⇒ [tex]=1.5\times 10^{-3}[/tex]
hence,
⇒ [tex]Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}[/tex]
On substituting the values, we get
⇒ [tex]1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}[/tex]
⇒ [tex]Mass \ of \ acid=1.5\times 10^{-3}\times 400[/tex]
⇒ [tex]=0.60 \ g[/tex]
