Question

How many grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel?

Answer 1

Answer: 0.87 g of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel.

Explanation:

According to the dilution law,

$C_1w_1+C_2w_2=C_3w_3$

where,

$C_1$ = concentration of ist progesterone gel = 8%w/w

$w_1$ = weight of ist progesterone gel  = x g

$C_2$ = concentration of another progesterone gel  = 4% w/w

$w_2$ = weight of another progesterone gel = 1.45 g

$C_3$ = concentration of resulting progesterone gel  = 5.5%w/w

$w_3$ = weight of resulting progesterone gel  = (x+1.45) g

$8\times x+4\times 1.45=5.5\times (x+1.45)$

$x=0.87$

Thus 0.87 g of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel.