How many grams of KBr are required to make 350. mL of a 0.115 M KBr solution?

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Dejanras Dejanras · Answer 1 · 2018-11-06T09:34:55+01:00

Answer is: 4.79 grams og KBr are required.

V(KBr) = 350 mL ÷ 1000 mL/L.

V(KBr) = 0.350 L; volume of potassium bromide solution.

c(KBr) = 0.115 M; molarity of potassium bromide solution.

n(KBr) = c(KBr) · V(KBr).

n(KBr) = 0.115 mol/L · 0.350 L.

n(KBr) = 0.04025 mol, amount of substance.

m(KBr) = n(KBr) · M(KBr).

m(KBr) = 0.04025 mol · 119 g/mol.

m(KBr) = 4.79 g; mass of potassium bromide.

M - molar mass.

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samueladesida43 samueladesida43 · Answer 2 · 2022-02-15T11:32:16+01:00

4.79 grams of KBr are required to make 350 mL of a 0.115M KBr solution.

The molarity of a solution can be calculated by using the following expression:

Molarity = no. of moles ÷ volume

According to this question, 350mL of a 0.115M KBr solution was given. The number of moles can be calculated as follows:

no. of moles = 0.350 × 0.115

no. of moles = 0.0403mol

Molar mass of KBr = 39 + 80 = 119g/mol

mass of KBr = 119g/mol × 0.0403mol

mass of KBr = 4.79g

Therefore, 4.79 grams of KBr are required to make 350 mL of a 0.115M KBr solution.

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