Solution Concentration

Question

(1)To make 1 liter of a 0.47 molar solution of sodium nitrate,NaNO^3, you add 0.47 moles to a volumetric flask. How many grams of NaNO^3 should you add? How much water should you add?

(2)To make 1000 grams of a 0.47% solution of sodium nitrate,NaNo^3 to a flask. How much water should you add?

(3)an experiment you need 2.0 L of 0.05 M hydrochloric acid HCL.You have 12 M hydrochloric acid stock solution. Explain the exact steps to prepare what you need.

According to the label , there is 1mL of medicine in the vial, and the concentration is 0.4mg/mL. If the doctor wants the patient to receive 0.2 mg, how many mL should the nurse give.

Balance eqaution

------Na^3 PO^4+------CaBr^2 arrow ------NaBr+------Ca ^3(PO^4)^2

------Al + ----H^2SO^4 arrow----Al^2(SO^4)^3+ ------ H^2

Answer 1

This looks a lot like a homework assignment. It is against Wyzant policy for tutors to do homework assignments. I will assist, but you need to try them and then if having difficulty, post what you've done and where you are having problems.

(1). 0.47 molar (M) means 0.47 moles of NaNO3 in 1 liter of solution.

How many grams? Find molar mass of NaNO3 using periodic table. That will be 1 mole. Multiply by 0.47 to get grams. The amount of water should be enough to make 1 liter of solution.

(2). 0.47% solution means 0.47 g per 100 g of solution. Since you have 1000 g of solution, you can find how much water (assuming each ml of water = 1 g).

(3). Dilution problem: use V1M1 = V2M2 where V1 and V2 are initial and final volumes, respectively and M2 and M2 are initial and final molarity, respectively.

(x L)(12 M) = (2.0 L )(0.05 M) and x = 0.008333 L = 8.3 mls

Take 8.3 mls of the 12 M HCl and add enough water to make a final volume of 2.0 L (2000 mls)

(4). 1 ml medicine is 0.4 mg / ml. If you want 0.2 mg how many mls?

If 1 ml has 0.4 mg, how many mls would it take to get 0.2 mg? Well ... 0.5 mls

(5). 2Na₃PO₄ + 3CaBr₂ ==> 6NaBr + Ca₃(PO₄)₂ (you wrote it with superscripts instead of subscripts)

2Al + 3H₂SO₄ ==> Al₂(SO₄)₃ + 3H₂ (you write it with superscripts instead of subscripts. Be careful)