50.0 ml sample of 0.108 m h2so4 is diluted to 250.0 ml. New molarity will be 0.216 M.
Every solute component present in a solution sample has lower concentration values due to dilution. The initial solution has been referred to as a stock solution, and water is typically used as the solvent to create an aqueous solution.
A sample of both the stock solution is made into a diluted version by adding extra pure liquid water. If a molarity is used to describe the solute concentration,
Given data:
[tex]C_{1} = 0.108 M\\C_{2} = 50mL\\V_{2} = 250mL[/tex]
New molarity can be determined by using the formula:
[tex]C_{1} V_{1} = C_{2} V_{2}\\C_{2} = C_{1} V_{1} / V_{2}\\[/tex]
[tex]C_{2}[/tex] = 0.108 M × 50 mL / 250 mL
[tex]C_{2}[/tex] = 0.216 M
New molarity of the of the sample will be 0.216 M.
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