# Can someone explain this question in detail? [closed]

An infusion stated to contain 0.95% w/v NaCl was diluted so that its Na content could be determined by flame photometry. The following dilutions were carried out:

1. 10 ml of the sample was diluted to 250 ml with water.
2. 10 ml of the diluted sample was diluted to 200 ml with water. The sample was found to contain 0.74 mg/100 ml of Na. Atomic weights: Na = 23 Cl = 35.5 Calculate: The % w/v of NaCl present in the infusion The % of stated content of NaCl. The concentration in mmoles/litre of NaCl present in the infusion.
• We will not make all the calculations without you trying to enter the problem. So try and start. Tell us where you find a difficulty. We will help you go ahead. Jun 16 '20 at 15:25
• It's an example question that I have the solution to and I just want it explained better.
– Amy
Jun 16 '20 at 15:36
• Better explaining without knowledge what has to be explained better is guesswork. Say explicitly in the question what you need to explain. Jun 16 '20 at 16:09
• It was this bit of the answer: Conc. of NaCl in the injection = 370 x (58.5/23) = 941 mg/100 ml
– Amy
Jun 16 '20 at 16:12
• Since you have an answer, would you answer my question?: Is infusion contains $0.94\% (w/v)$? Jun 16 '20 at 16:30

In flame photometry, we need to dilute the NaCl sample because the calibration is not linear at high NaCl concentration.

It is a classical dilution problem. I find it useful to remember

$$C_iV_i=C_fV_f$$

where $$i$$ indicates initial concentration (before dilution) and volume and $$f$$ indicates final concentration and volume (after dilution).

Now note that your sample concentration is in 0.95% w/v NaCl. The flame photometer only responds to Na ions in the solution. So your flame photometer results tells you how much Na is there, not NaCl.

Try to use mole ratios between

grams of Na => moles of Na

What is the relationship between moles of Na and moles of NaCl?