Question

How many grams of chromium(II) nitrate, , must be dissolved to prepare 500. mL of a 0.188 M aqueous solution of the salt?

Answer 1

22.37 g of chromium(II) nitrate, must be dissolved to prepare 500. mL of a 0.188 M aqueous solution of the salt.

According to the definition, the molar concentration of a substance in a solution is the ratio of the number of the moles to the volume of the solution:

c=n/V.

The number of the moles is related to the mass with the molar mass:

n=m/M;

m=n·M.

Thus, given the volume of the solution of chromium(II) nitrate, its concentration and molar mass is 238.011 g/mol we can calculate the mass of chromium(II) nitrate needed for the preparation :

∴   Cr(NO₃)₃ = cVM

                 = 0.188 M × 0.5 L × 238.011 g/mol

                 = 22.37 g

Therefore, 22.37 g of chromium(II) nitrate, must be dissolved to prepare 500. mL of a 0.188 M aqueous solution of the salt.

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