Q5. Two beakers, A and B, each contain a 100 ml solution of sodium hydroxide and water. Beaker A contains 35 g of sodium hydroxide, and Beaker B contains 50 g of sodium hydroxide. In which beaker is the base solution more dilute? Explain.
Answer (1)
To determine which beaker contains the more dilute solution, we need to compare the concentration of sodium hydroxide (NaOH) in each beaker. Concentration is typically expressed in terms of mass per unit volume, which in this case can be calculated as grams of NaOH per 100 mL of solution.
Beaker A:
Contains 35 g of NaOH in 100 mL of solution.
The concentration of NaOH in Beaker A is: 100 mL 35 g NaOH = 0.35 g/mL
Beaker B:
Contains 50 g of NaOH in 100 mL of solution.
The concentration of NaOH in Beaker B is: 100 mL 50 g NaOH = 0.50 g/mL
By comparing these concentrations, we can see that Beaker A has a concentration of 0.35 g/mL, while Beaker B has a concentration of 0.50 g/mL. Therefore, Beaker A has the lower concentration of sodium hydroxide, making it the more dilute solution.
In summary, Beaker A contains the more dilute sodium hydroxide solution because it has a lower concentration of NaOH.
