Question

A sample 25.0 mL of an NaCl solution is diluted to a final volume of 125.0 mL of 0.200 M NaCl.What was concentration of the original NaCl solution?How many moles of solute are pipetted from the original solution?How many moles of solute are in the dilute solution?

How many moles of product, KHCO₃ will be produced when1.0 mol of K₂O reacts with 1.0 mol H₂O and 1.0 mol of CO₂?K₂O + H₂O + 2 CO₂ à 2 KHCO₃

h) A metal forms +2 ions. Its bromide compound has a molar mass of 297.135 grams/mol. Identify the metal.

i) What volume in liters of 0.556 M KCl is needed to obtain 3.10 grams of KCl?

j) Which of the following is a non-electrolyte? NaOH, HCl, KCl, C₆H₁₄

Answer 1

1) Using N1V1 = N2V2

we can calculate concentrations of the 25 mL sample

125 x 0.2 = 25 x N2

So the original solution was a 1M solution.

Number of moles of solute that was pipetted out from the original solution was

= 1M x 25 mL/1000 = 0.025 moles

The dilute solution also has 0.025 moles of the solute.

2) K₂O + H₂O + 2CO₂ 2KHCO₃

1.0 mol of K₂O reacts with 1.0 mol H₂O and 1.0 mol of CO₂

As per the stoichiometry of the reaction the limiting reagent is CO₂ since we are taking 1.0 mole

which should give 1.0 mole of KHCO₃

3) A metal forms +2 ions. Its bromide compound has a molar mass of 297.135 grams/mol. Identify the metal.

If the metal is +2 charge it will take two Bromine atoms so the mass from bromine will be 160. The remaining mass is 137. So the element is Barium.

4)  What volume in liters of 0.556 M KCl is needed to obtain 3.10 grams of KCl?

1mole of any compound in 1000 mL is 1M

MW of KCl is 74.55, 3.1 g will be 0.04158 moles.

Concentration x volume in L = moles

0.556 M x volume =0.04158

volume = 0.04158/0.556 = 0.07478 L that is 74.78 mL

5) j) Which of the following is a non-electrolyte? NaOH, HCl, KCl, C6H14

C6H14 is a non-electrolyte.