Solve the following problems. Show all your work. 1. 2. E. Calculate the molarit

Question

Solve the following problems. Show all your work. 1. 2. E. Calculate the molarity (M) of 30.0 g of NaOH in 350 mL of NaOH solution. Calculate the amount of solute, in moles and in grams, needed to prepare 1.00 L of a 3.00 M NaCl solution. Calculate the amount of solute in grams needed to prepare 2.00 L of a 1.50 M NaOH solution. 3. 2.0 L of a 6.0 M HCl solution is added to water so that the final volume is 6.0 L. Calculate the new concentration 4. What is the concentration of K,CO,, in molarity, of a solution made from 5.00 grams of potassium carbonate mixed into 250 mL 5. of water? e you have 125.0 mL of 0.400 M NaCl solution. This is too salty, so 6. Suppos ou dilute it by adding another 50.0 mL of water: What is the new concentra- tion of NaCl in units of molarity? 200 mL of a 0.500 M Ca(NO,)2 solution was made, but only 150 mL of solu- tion are needed. If you pour out 50 mL of the solution, what is the concentra- tion, in units of molarity, of the 150 mL of solution you have left? 7.

Explanation / Answer

1. Molarity = no.of moles of solute/ volume of solution in litres

Molar mass of NaOH = 40gm/mol

no.of moles of NaOH in 30 gm = 30/40 = 0.75

Volume of solution = 350 ml = 0.35 litres

molarity = 0.75/0.35 = 2.142 M

2. Molar mass of NaCl = 58.5gm/mol

volume of solution = 1 litre

Molarity of solution = 3M

Moles of solute = molarity x volume = 3 x 1 = 3

mass of NaCl = moles of NaCl x molar mass of NaCl = 3 x 58.5 = 175.5 gm

3.

Molar mass of NaOH = 40 gm/mol

volume of solution = 2 litre

Molarity of solution = 1.5M

Moles of solute = molarity x volume = 1.5 x 2 = 3

mass of NaOH = moles of NaOH x molar mass of NaOH = 3 x 40 = 120 gm

4. Volume of HCl solution = 2 litre

Initial molarity of HCl = 6M

Number of moles of HCl = volume x molarity = 2 x 6 = 12 moles

Final volume = 6 litre

So, new molarity = moles of HCl/final volume = 12/6 = 2 M

5. mass of potassium carbonate = 5 gm

volume of water = 250 ml = 0.25 litre

molar mass of potassium carbonate = 138 gm/mol

moles of potassium carbonate = 5/138 = 0.0362

molarity = moles of potassium carbonate/volume = 0.0362/0.25= 0.1449 M

6. Molarity of NaCl solution = 0.4M

Volume of NaCl solution = 125 ml = 0.125 litre

Moles of NaCl = molarity x volume = 0.125 x 0.4 = 0.05 moles

Final volume = initial volume + water added = 125+50=175 ml = 0.175 litre

New molarity = moles of NaCl/final volume = 0.05/0.175 = 0.2857M

7. Volume of Calcium nitrate solution = 200 ml

Molarity = 0.5M

Since a solution has homogeneous concentration, so a part of solution will have concentration same as that of whole solution.

Thus, concentration of 50 ml solution will be same as concentration of 150 ml solution which will be same as concentration of whole solution of 200 ml = 0.5M

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