ld tul using une assumption above. Estimated volume of an Elodea cell: cuiate Co

Question

ld tul using une assumption above. Estimated volume of an Elodea cell: cuiate Complete the following conversions: ot 0.72 kg to mg 332 to g 2.25 m to cm 0.435 L to mL You are asked to make a 1:10 dilution of a 3.5% NaCl stock solution to a total volume of 10 ml. How much stock and how much water are needed: stock: Convert this 3.5% solution to grml. Given a MW of 58.44 g/mole. Convert the ug/ml above to moles/L (molar concentration) mi water Skills: Prgper and safe hanaampound light microscope and know their functions handling of a microscope.

Explanation / Answer

1kg =1000gms

1gms = 1000mg.

So 1 kg=1000000mg.

So 0.72kg =720000mg.

1gm =1000mg

1mg= 1000microgram.

So 322 microgram=3.22*10^-4gms

1m =100cms

So 2.25m =225 cms

1litre =1000ml.

So 0.435l= 435ml

Stock 1ml

Water 9ml.

So final percentage of NaCl is 0.35%.

Means 0.35 gms of NaCl in 100ml of water

350 mg in 100ml water or 3.5mg in 1ml

or 3500 microgram/ml

So mole per liter will be

3500microgram = 3.5*10-3 gms

3.5 * 10-3/58.55 * 1000/1

58.9 micromol.

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