Question

A prescription balance has a sensitivity requirement of 0.003 gram. Explain how you would weigh 0.008 gram of atropine sulfate with an error not greater than 5% using lactose as diluent.

Answer 1

( i ) smallest quantity may be weighed = 100% x sensitivity requirement / acceptable error (%)

= 100% x 0.003 g/ 5 % = 0.06 g

( ii ) 0.008 g of atropine sulphate needed which is right times less than smallest quantity to be weighed

Multiplying factor 8

= 0.06 gbof drug and 0.06 x 8 will be drug + diluents = 0.48 g

( iii ) prepare a mixture of total 0.48 g

Include 0.06 g of drug and 0.42 g diluent

( iv ) Now mixture of 0.48 is divided into 8 aliquot

Each part will have = 0.006 g drug g drug ( atropine sulphate) and 0.042 g of diluent ( lactose )

Thankyou