A. In the laboratory a student dilutes 19.1 ml of a 11.2 M nitric acid solution

Question

A. In the laboratory a student dilutes 19.1 ml of a 11.2 M nitric acid solution to a total volume of 100.0 mL. What is the concentration of the diluted solution?
B. How many mililiters of 6.83 M nitric acid solution should be used to prepare 3.50 L of 0.100 M HNO3?
C. In the laboratory a student adds 54.0 mL of water to 21.0 mL of a .694 M hydrochloric acid solution. What is the concentration of the diluted solution? Assume values are additive A. In the laboratory a student dilutes 19.1 ml of a 11.2 M nitric acid solution to a total volume of 100.0 mL. What is the concentration of the diluted solution?
B. How many mililiters of 6.83 M nitric acid solution should be used to prepare 3.50 L of 0.100 M HNO3?
C. In the laboratory a student adds 54.0 mL of water to 21.0 mL of a .694 M hydrochloric acid solution. What is the concentration of the diluted solution? Assume values are additive
B. How many mililiters of 6.83 M nitric acid solution should be used to prepare 3.50 L of 0.100 M HNO3?
C. In the laboratory a student adds 54.0 mL of water to 21.0 mL of a .694 M hydrochloric acid solution. What is the concentration of the diluted solution? Assume values are additive

Explanation / Answer

A) M1V1 = M2V2

M1 = 11.2 M , V1 = 19.1 mL

M2 = ? , V = 100.0 mL

M2 = (M1V1 / V2)

M2 = (11.2 x 19.1 / 100)

M2 = 2.14 M

B) M1V1 = M2V2

M1 = 6.83 M , V1 = ?

M2 = 0.100 M , V2 = 3.50 L = 3500 mL

V1 = (M2V2 / M1)

V1 = (0.100 x 3500 / 6.83)

V1 = 51.24 mL

C) M1V1 = M2V2

M1 = 0.694 M , V1 = 21.0 mL

M2 = ? , V2 = 54.0 + 21.0 = 75.0 mL

M2 = (M1V1 / V2)

M2 = (0.694 x 21 / 75)

M2 = 0.1943 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.