# Chemistry

1. You have been given a sample of unknown molarity. Calculate the molarity of a solution which has been prepared by dissolving 8.75 moles of sodium chloride in enough water to produce a solution of 6.22l.

2. You have a sample which consists of 428g sodium hydroxide (NaOH) dissolved in enough water to make a final solution of 6.4l. What is the molarity?

3. You have been given a sample of 4.85 moles of glucose (C6H12O6) and you want to make a 0.75M solution with it. What will be the final volume of solution?

4. 37.0cm³ of a sodium hydroxide (NaOH) solution was pipetted into a conical flask and titrated with a standard solution of 0.45 mol dm-3 (0.3M) hydrochloric acid according to the following equation:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

A universal indicator solution was used for the titration and it was found that 22.0 cm3 of the acid (HCl) was required to neutralise the alkali (NaOH). Calculate the molarity of the sodium hydroxide and its concentration in g/dm3.

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1. The first three questions may be solved using the relation:
molarity (mols)
= number of moles / volume (L)
where
number of moles
= mass (g) / molar mass (g).

Example:
10.0 g of NaCl is dissolved to make 0.523 L of solution. Find molarity of the NaCl solution.
number of moles
= 10/(23+35.5)=0.171
molarity
=0.171/0.523
=0.327

For #4,
From the equation, it can be deduced that neutralization requires equal moles of acid and base.
=>
#of moles used for acid and base are equal.
Use the above relation to solve for # of moles.

Note: the molarity of HCl is not clear. 0.45 mol-dm-3≡0.45 mol/L but does NOT equal 0.3M.

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