calculating molarity?

Question

if a sample of 8.26 g of Barium hydroxide dissolved and diluted to the mark in a 250 mL flask.
it was found that 8.69 mL of this solution was needed to reach the stoichimetric point in a titration of 28.6 mL of HNO3. What is the molarity of the HNO3 solution?

i did it out and got 3.37 M?
is this correct?

Answer 1

There are 250mL volumetric flasks in which the mark on the neck indicates exactly 250mL if you fill it to that mark with the bottom of the meniscus touching the mark .So then!

Atomic weights: Ba = 137 O =16 H = 1 Ba(OH)2 = 171

Ba(OH)2 + 2HNO3 ===> Ba(NO3)2 + 2H2O

Yours is a a two-part solution, moles/L or moles/1000mL, so you need a two-part beginning:

8.26gBa(OH)2/250mLBa((OH)2soln x 1000mL/1L x 1molBa(OH)2/171gBa(OH)2 x 2molHNO3/1molBa(OH)2 = (8.26)(1000)(2)/(250)(171) =