What is the molarity of a solution prepared with 25 g NaCl in 650 ml solution? m

Question

What is the molarity of a solution prepared with 25 g NaCl in 650 ml solution? mass 01 Nacl = 2 kg molar mass of uaCl = 58.5 g 1 mol if you were in the laboratory, how many ml of a 1200 M naOh stock solution would you need to prepare 750.0 ml of 0.200 m NaOh? 750 ml times 1l/1000ml = 750/1000 = 0.75l + Na- 2t 900 0-16 H - 1.0079 40 0.75 How many l of a 1.5 M glucose solution will contain 3, 0 moles? 1.5 m + 3.0 moles = 4.50L What is the concentration of a solution formed by dilution 125 ml of 12.0 M HCL solution to 850ml?

Explanation / Answer

Solution :-

Q2) Given data

Mass of NaCl = 25 g

Volume = 650 ml = 0.650 L

Moles of NaCl = mass of NaCl / molar mass of NaCl

                          = 25 g / 58.443 g per mol

                          = 0.428 mol

Molarity = moles/ volume in liter

                 = 0.428 mol / 0.650 L

                  = 0.66 M

Q3) Given data

Initial volume = ?

Initial molarity = 12.00 M

Final volume = 750.0 ml

Final molarity = 0.200 M

M1V1 = M2V2

V1 = M2V2/M1

V1= 0.200 M * 750.0 ml / 12.00 M

V1 = 12.50 ml

So the volume needed is 12.50 ml

Q4)Given data

Molarity = 1.5 M

Moles = 3.0 moles

Volume = ?

Molarity = moles / liter

liter = moles / molarity

liter = 3.0 moles / 1.5 mol per L

liter = 2.0 L

So the volume of glucose needed is 2.0 L

Q5) Given data

Initial volume = 125 ml

Initial molarity = 12.0 M

Final volume = 850 ml

Final molarity = ?

M1V1= M2V2

M2= M1V1/V2

M2 = 12.00 M* 125 ml / 850 ml

M2= 1.76 M

So the molarity of the diluted solution is 1.76 M

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