49.7 ml of a solution of naoh is needed to neutralize a solution that contains 1.86 g of h2504 in 20.00 ml of water. what is the molarity of the naoh solution?
h2so4 + 2 naoh --> 2h2o + na2so4

1) 0.763 m
2) 0.0404 m
3) 1.13 m
4) 0.481 m
5) 0.602 m
6) 2.23 m
7) 1.81 m
8) 2.94 m

1) C NaOH = 0.763 M

Explanation:

H2SO4 + 2 NaOH → 2 H2O + Na2SO4

∴ V NaOH = 49.7 mL = 0.0497 L

∴ m H2SO4 = 1.86 g

∴ V sln = 20.00 ml = 0.020 L

∴ Mw H2SO4 = 98.079 g/mol

⇒ n H2SO4 = (1.86 g)(mol / 98.079 g) = 0.019 mol H2SO4

⇒ n NaOH = (0.019 mol H2SO4)(2 mol NaOH/ mol H2SO4) = 0.038 mol NaOH

⇒ C NaOH = 0.038 mol / 0.0497 L = 0.763 M

U whay do you need with?

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.