You record a %t of 85 from a spec 20.
this is equivalent to what absorbable level?

0.096 M

General Formulas and Concepts:

Chemistry - Solutions

Explanation:

Step 1: Define

6.25 g Ba(NO₃)₂

250 mL

Step 2: Define conversions

1000 mL = 1 L

Molar Mass of Ba - 137.33 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Ba(NO₃)₂ - 137.33 + 2(14.01) + 6(16.00) = 261.35 g/mol

Step 3: Convert

= 0.25 L

= 0.023914 mol Ba(NO₃)₂

Step 4: Find Molarity

M = 0.023914 mol Ba(NO₃)₂ / 0.25 L

M = 0.095657

Step 5: Check

We are given 2 sig figs as our smallest. Follow sig fig rules.

0.095657 M ≈ 0.096 M