Answer to Question #126007 in Physical Chemistry for asad

Question #126007
How many grams of KCl are in 225 g of an 8% (m/m) solution?

How many grams of NaHCO3 is in 325 mL of 4.5 M solution of NaHCO3?

What is the molarity of a solution that contains 75 g of KNO3 in 350 mL of solution?

How many mL of a 0.3 M glucose (C6H12O6) intravenous solution is needed to deliver?

What volume (mL) of 0.2 M HCl solution can be prepared by diluting 50 mL of 1 M HCl?

How many grams of solute are needed to prepare 150 mL of a 40% (m/v) solution of LiNO3?

What is the mass % (m/m) of a solution prepared by dissolving 30.0 g of NaOH in 120.0 g of water?

What is the mass % (m/v) of a solution prepared by dissolving 5.0 g of KI to give a final volume of 250 mL?

A topical antibiotic solution is 1.0% (m/v) Clindamycin. How many grams of Clindamycin is in 65 mL of this solution?
1
Expert's answer
2020-07-13T07:28:43-0400

Solution.

1.

"w = \\frac{m(KCl)}{m(solution)}\\times 100\\%"

"m(KCl) = \\frac{w \\times m(solution)}{100\\%}"

m(KCl) = 18.0 g

2.

"C = \\frac{m}{M \\times V}"

"m = C \\times M \\times V"

m(NaHCO3) = 122.86 g

3.

"C = \\frac{m}{M \\times V}"

C(KNO3) = 2.12 M

4.

"C = \\frac{m}{M \\times V}"

"V = \\frac{m}{C \\times M}"

V(C6H12O6) = 0.185 L = 185.0 ml

5.

"C1 \\times V1 = C2 \\times V2"

"V2 = \\frac{C1 \\times V1}{C2}"

V2 = 250.0 ml

6.

"\\rho(LiNO3) = \\frac{m(LiNO3)}{V(Solution)}\\times100\\%"

"m(LiNO3) = \\frac{\\rho(LiNO3) \\times V(Solution)}{100\\%}"

m(LiNO3) = 60 g

7.

"w = \\frac{m(NaOH)}{m(NaOH) + m(H2O)}\\times100\\%"

w = 20.0 %

8.

"\\rho(KI) = \\frac{m(KI)}{V(solution)}\\times100\\%"

"\\rho(KI) = 2.0 \\%"

9.

"\\rho(antib.) = \\frac{m(antib.)}{V(solution)}\\times100\\%"

"m(antib.) = \\frac{\\rho(antib.) \\times V(solution)}{100\\%}"

m(antib.) = 0.65 g

Answer:

1.

m(KCl) = 18.0 g

2.

m(NaHCO3) = 122.86 g

3.

C(KNO3) = 2.12 M

4.

V(C6H12O6) = 0.185 L = 185.0 ml

5.

V2 = 250.0 ml

6.

m(LiNO3) = 60.0 g

7.

w = 20.0 %

8.

"\\rho(KI) = 2.0 \\%"

9.

m(antib.) = 0.65 g


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