Answer to Question #236515 in General Chemistry for Russ

Question

Prepare 50ml of 0.3N Ba(OH)2 * 8 H2O

Answer 1

Normality = $\frac{no.of.equivalent×1000}{volume.of.solution(mL)}$

0.3 = $\frac{no.of.equivalent×1000}{50}$

no. of equivalent = 0.015

no. of equivalent = no. of moles/ n factor

0.015 = no. of moles / 2

no. of moles = 0.030

( Gram required / molar mass )= 0.030

Gram required = 0.030 × 315.46

Gram required = 9.46 g

9.46 g of Ba(OH)₂ is required to prepare 0.3N solution