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Average tablet weight = 12.7592/20 = 0.63796g
0.1733g in total 200ml - take 10ml filtrate, so would contain (8.665x10^-3)g, and make up to 100ml. Take another 10ml filtrate and make up to 100ml.
Now you have (8.665x10^-4)g in 100ml.
Beer lambert law: A = A(1%,1cm) x path length x conc(g/100ml)
0.540 = 715 x 1 x conc
Therefore conc = (7.55 x 10^-4)g/100ml.
This is conc of paracetamol ( maybe) in 100ml, and you have powdered tablet mix conc of (8.665 x10^-4)g, so I think answer is
(7.55x10^-4)/(8.665x10^-4) x 0.63796 = 0.555868205g = 556mg
Again I'm not 100% on the answer but i think this is it
Noscopine Linctus (16.9249 g) was diluted to 100 ml with 0.05M HCl and 3 ml of this solution was then diluted to 50 ml with water. A 2-cm layer of the resulting solution was found to have an absorbance of 0.561 at the maximum at 310 nm.
The weight per ml of the linctus was 1.229 g ml-1.
Taking 90.7 as the value of A(1%, 1cm), calculate the mass (mg) of Noscopine in a 5 ml dose of the linctus.
similar qu but what confuses me is do you find mass in 3ml first or 50ml, if its 3ml then surely mass would remain same in 50ml.
Yes you find mass in 3ml and yes it's the same if you make up to 50ml because you're not adding any drug. However you need to take into account the resulting dilution factor in subsequent calculations.
20 Paracetamol Tablets (total weight 12.7592 g) were crushed and powdered. 0.1733 g of the powder was added to 50 ml of 0.1M sodium hydroxide, diluted with 100 ml of water, shaken for 15 minutes and then sufficient water to produce 200 ml was added. The resulting solution was filtered and 10 ml of the filtrate was diluted to 100 ml with water. 10 ml of the resulting solution was added to 10 ml of 0.1M sodium hydroxide followed by dilution to 100 ml with water. A 1-cm layer of the resulting solution had an absorbance of 0.540 at the maximum at 257 nm. Calculate the mass (mg) of paracetamol in a tablet of average weight, taking 715 as the value of A(1%, 1cm) at the maximum at 257 nm.
Give the result correct to the nearest mg.
Would the oringial dilution be:
0.540/715 * 100/10 * 100/10 or have I missed something out as I keep getting the wrong answer.
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