The millimoles of HCl needed to neutralize per gram of sample is [tex]\boxed{{\text{0}}{\text{.0037 millimoles}}}[/tex].
Further explanation:
Molarity:
The molarity of the solution is defined as the concentration of the solution. It is the number of moles of the solute dissolved in one liter of the solution.
The formula to relate molarity (M), volume (V), and number of moles (n) is as follows:
[tex]{\text{M}} =\dfrac{{\text{n}}}{{\text{V}}}[/tex] …… (1)
Here V is a volume of solution, n is a number of moles of solute, and M is the molarity.
On rearranging equation (1) for n. We get,
[tex]{\text{n}} = {\text{M}}\times {\text{V}}[/tex] …… (2)
Given molarity of HCl is 0.4548 M.
Volume of HCl is 14.10 mL.
As the volume is in mL, the unit of moles will be millimoles.
Substitute these values in equation (2) to calculate the initial moles of HCl.
[tex]\begin{aligned}{\text{n}}&= {\text{0}}{\text{.4548 M}} \times {\text{14}}{\text{.10 mL}}\\&={\text{6}}{\text{.41268 millimoles}}\\\end{aligned}[/tex]
The number of moles of HCl is 6.41268 millimoles.
Given molarity of NaCl is 0.1048 M.
Volume of NaCl is 14.60 mL.
Substitute these values in equation (2) to calculate moles of NaCl.
[tex]\begin{aligned}{\text{n}}&= {\text{0}}{\text{.1048 M}} \times {\text{14}}{\text{.60 mL}}\\&={\text{1}}{\text{.53008 millimoles}}\\\end{aligned}[/tex]
The number of moles of NaCl formed is 1.53008 millimoles.
The moles of HCl reacted to neutralize the tablet are calculated as follows:
[tex]\begin{aligned}{\text{Moles of HCl}}\left( {{\text{reacted}}} \right) &= {\text{Initial}}\;{\text{moles}} - {\text{moles of}}\;{\text{NaCl}}\\&= {\text{6}}{\text{.41268 millimoles}} - 1.53008\;{\text{millimoles}}\\&= {\text{4}}{\text{.8826 millimoles}}\\\end{aligned}[/tex]
The moles of HCl reacted to neutralize the tablet are 4.8826 millimoles.
The conversion factor to convert mass in mg to g is as follows:
[tex]{\text{1g}} = 1000\;{\text{mg}}[/tex]
The mass of tablet is 1.333 g.
The mass of tablet in mg is calculated as follows:
[tex]\begin{aligned}{\text{Mass}}&= \,\left( {{\text{1}}{\text{.333}}\;{\text{g}}}\right)\left( {\frac{{{\text{1000}}\;{\text{mg}}}}{{{\text{1}}\;{\text{g}}}}} \right)\\&= {\text{1333}}\;{\text{mg}}\\\end{aligned}[/tex]
4.8826 millimoles of HCl neutralize 1333 mg of tablet. Therefore, millimoles of HCl required to neutralize per gram of tablet are calculated as follows:
[tex]\begin{aligned}{\text{Moles}}&= \left( {1\;{\text{mg}}} \right)\left( {\frac{{4.8826\;{\text{millimoles}}}}{{1333\;{\text{mg}}}}} \right)\\&= 0.003662\,{\text{millimoles}}\\&= 0.0037\;{\text{millimoles}}\\\end{aligned}[/tex]
Hence, 0.0037 millimoles of HCl are needed to neutralize per gram of tablet.
Learn more:
1. Determine the coefficients needed to balance the equation for the complete combustion of methane: https://brainly.com/question/1971314
2. Determine the energy required to break the hydrogen bond: https://brainly.com/question/10278354
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Solutions
Keywords: molarity, millimoles, solution, number of moles, HCl, NaCl, tablet, 1333g, 14.10 mL, 14,60 mL and 0.003226 millimoles.
