Question

1)  What g mass (two decimal places) of 9 % w/w of hydrocortisone ointment is needed to be mixed with 65 g of 2.0 % w/w hydrocortisone ointment to make 3.7 % w/w hydrocortisone ointment?

2) How many mg (round to a whole number) of sodium fluoride are needed to prepare 250 mL of a sodium fluoride stock solution such that a solution containing 1.0 ppm of sodium fluoride results when 1.2 mL is diluted to 3800 mL?

3) What g mass (two decimal places) of 9 % w/w of hydrocortisone ointment is needed to be mixed with 65 g of 2.0 % w/w hydrocortisone ointment to make 3.7 % w/w hydrocortisone ointment?

Please help!!

Answer 1

1)

Let W g of 9 % w/w of hydrocortisone ointment is mixed with 65 g of 2.0 % w/w hydrocortisone ointment

Weight of hydrocortisone in W g of 9 % w/w of hydrocortisone ointment

... ...(1)

W9 X 100

Weight of hydrocortisone in 65 g of 2.0 % w/w of hydrocortisone ointment

... ...(2)

= 65 g x 120 100

Final weight percent of hydrocortisone ointment is 3.7%.

Weight of hydrocortisone in of 3.7 % w/w of hydrocortisone ointment

(W + 65 g

$=\left [ \left ( W+65 \right ) \ g \times \frac{3.7}{100} \right ]$ ... ...(3)

But

(1) + (2) = (3)

$\left [ W \ g \times \frac{9}{100} \right ] + \left [ 65 \ g \times \frac{2}{100} \right ]=\left [ \left ( W+65 \right ) \ g \times \frac{3.7}{100} \right ]$

0.09W + 1.3 = 0.037(W + 65)

0.09W +1.3 = 0.037W + 2.405

0.09W -0.037W = 2.405 – 1.3