Answer to Question #126007 in Physical Chemistry for asad

Question

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?

Answer 1

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