In the laboratory you dissolve 211 g of chromiumII nitrate i

In the laboratory you dissolve 21.1 g of chromium(II) nitrate in a volumetric flask and add water to a total volume of 375 What is the molarity of the solution? What is the concentration of the chromium(II) cation? What is the concentration of the nitrate anion? mL. M. M. M. Submit Answer Retry Entire Group 4 more group attempts remaining Next (Pre

Solution

Chromium(II)nitrate = Cr(NO3)2

Molar mass of Cr(NO3)2 = 238 g/mol
So, 238 g of Cr(NO3)2 = 1 mol
or, 1 g of Cr(NO3)2 = (1/238) mol
or, 21 g of Cr(NO3)2 = (21/238) mol = 0.088 mol

Volume = 375 mL = 0.375 L

[Cr(NO3)2] = 0.088 mol / 0.375 L = 0.235 M

Now,

Cr(NO₃)₂    <--------->    Cr³⁺     +     2NO₃^-

1 mole of Cr(NO₃)₂ dissociate into 1 mole of Cr³⁺ and 2 moles of NO₃^-

So,
[Cr³⁺] = 0.235 M
[NO₃^-] = 2 x 0.235 M = 0.470 M

(2)

Barium hydroxide = Ba(OH)2

Molar mass of Ba(OH)2 = 171.34 g/mol
So, 171.34 g of Ba(OH)2 = 1 mol
or, 1 g of Ba(OH)2 = (1/171.34) mol
or, 12.6 g of Ba(OH)2 = (12.6/171.34) mol = 0.074 mol

Volume = 375 mL = 0.375 L

[Ba(OH)2] = 0.074 mol / 0.375 L = 0.197 M

Now,

Ba(OH)₂    <--------->    Ba²⁺     +     2OH^-

1 mole of Ba(OH)₂ dissociate into 1 mole of Ba²⁺ and 2 moles of OH^-

So,
[Ba²⁺] = 0.197 M
[OH^-] = 2 x 0.197 M = 0.394 M