Answer is: 4.79 grams og KBr room required.

You are watching: How many grams of kbr are required to make 350. ml of a 0.115 m kbr solution?

V(KBr) = 350 mL ÷ 1000 mL/L.

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

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

n(KBr) = c(KBr) · V(KBr).

n(KBr) = 0.115 mol/L · 0.350 L.

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

m(KBr) = n(KBr) · M(KBr).

m(KBr) = 0.04025 mol · 119 g/mol.

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

M - molar mass.

of KBr is forced to do 350 mL of 0.115 M KBr solution.

Further Explanation:

Concentration is characterized as the lot or amount of solute existing in specific amount of solution. In stimulate to evaluate concentration of miscellaneous solutions, different concentration terms space used. Few of these are mentioned below.

1. Molarity (M)

2. Molality (m)

3. Mole portion (X)

4. Components per million (ppm)

5. Massive percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is defined as the number of moles of solute that can be liquified in one litre that the solution. The is denoted by M and also its unit is mol/L.

The formula to calculation molarity that KBr equipment is as follows:

...... (1)

Rearrange equation (1) to calculate moles that KBr.

ight>endaligned}" /> ...... (2)Substitute 0.115 M for molarity that KBr solution and also 350 mL because that volume the KBr equipment in equation (2).

The formula to calculation moles of KBr is as follows:

...... (3)Rearrange equation (3) for mass that KBr.

...... (4)Substitute 0.04025 mol because that moles of KBr and 119.002 g/mol because that molar mass of KBr in equation (4).

Therefore 4.79 g the KBr is required to make 350 mL of a 0.115 M KBr solution.

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Answer details:

Grade: senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, KBr, 350 mL, 0.115 M, 4.79 g, mass, molar mass, concentration terms, concentration, solute, moles of solute, volume.