Answer to Question #125896 in Physical Chemistry for Ahmad ali

Question

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 are in 65 mL of this solution?
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?

Answer 1

1.

Mass of KI 5.0 g .

final volume of Solution 250 ml .

% m /v of KI = Mass of KI in g . * 100 . / Volume of Solution in ml .

= 5 g * 100 / 250 ml ..

= 2 % ( m / v ) .

2 .

A topical antibiotic solution is 1.0% (m/v) Clindamycin .

Volume of Solution = 65 ml.

So , Mass of  Clindamycin = (65 * 1 / 100) g . = 0.65 g .

3 .

Molar Mass of KNO₃ = 39 + 14 + 48 . = 101 g / mol .

Moles Of KNO₃ = Mass of KNO₃ in g / Molar Mass of KNO₃ ..

= 75 g / 101 g /mol .

= 00.7425 Mol.

Volume of Solution = 350 ml . = 350 / 1000 = 0.350 L .

Molarity = Moles of KNO₃ / Volume of Solution in L .

= 0.7425 mol . / 0.350 L .

= 2.1214 M = 2.1214 mol / L .

4.

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

Generally 5 % solution is used

So , V = 500 ml ( approx ) .

CALCULATION FOR 500 ML ..............

If we Consider 500 ml Solution Then Moles Of glucose =( 500 * 0.3 ) / 1000 . .

= 0.15 Mol .

So , Mass Of Glucose = 0.15 * 180 = 27 g .

So , % = 27* 100 / 500 = 5. 4 %

So , answer 500 ml. ( APPROX .)