the teacher instructed three students A,B,and C respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). 'A' dissolved 50 g of NaOH in 100 ml of water B dissolved 50g of NaOH in 100g of water while C dissolved 50g of NaOH in water to make 100ml of solution . which one of them has made the desired solution and why ?

c has made the desired solution

To determine which student has made the desired solution, we need to calculate the mass by volume percentage of sodium hydroxide (NaOH) for each of them.

Student A dissolved 50 g of NaOH in 100 ml of water. To calculate the mass by volume percentage, we need to divide the mass of NaOH by the total volume of solution (NaOH + water) and multiply by 100.

Mass of NaOH in Student A's solution = 50 g
Volume of solution in Student A's solution = 100 ml + 100 ml (water) = 200 ml

Mass by volume percentage = (50 g / 200 ml) * 100% = 25%

Student B dissolved 50 g of NaOH in 100 g of water. Again, we need to divide the mass of NaOH by the total volume of solution (NaOH + water) and multiply by 100.

Mass of NaOH in Student B's solution = 50 g
Volume of solution in Student B's solution = 50 g (NaOH) + 100 g (water) = 150 g

Mass by volume percentage = (50 g / 150 g) * 100% ≈ 33.33%

Student C dissolved 50 g of NaOH in water to make 100 ml of solution. We need to calculate the mass by volume percentage using the same method as before.

Mass of NaOH in Student C's solution = 50 g
Volume of solution in Student C's solution = 100 ml

Mass by volume percentage = (50 g / 100 ml) * 100% = 50%

Out of the three students, Student C has made the desired solution of 50% (mass by volume) sodium hydroxide. This is because the mass by volume percentage of Student C's solution is closest to the desired percentage of 50%.