A patient received 5.0 g of NaCl in 8 hours. How many milliliters of a 0.40 % (m

Question

A patient received 5.0 g of NaCl in 8 hours.

How many milliliters of a 0.40 % (m/v) NaCl (saline) solution were delivered?

Express the volume in milliliters to two significant figures.

What is the molarity of the acetic acid solution if 30.7 mL of a 0.215 M KOH solution is required to titrate 29.0 mL of a solution of HC2H3O2?

HC2H3O2(aq)+ KOH(aq)? H2O(l)+ KC2H3O2(aq)

If 42.8 mL of a 0.156 M NaOH solution is required to titrate 15.5 mL of a solution of H2SO4, what is the molarity of the H2SO4 solution?

H2SO4(aq)+2NaOH(aq)?2H2O(l)+Na2SO4(aq)

Express your answer with the appropriate units.

Explanation / Answer

1 ans

Saline concentration is 0.4grams per ml .

We need 5 grams of NaCl . Hence required volume is 5/0.4 = 12.50ml

2 ans

From the stoichiometric equation , one mole of acetic acid reacts with one mole of potassium hydroxide . Number of moles of potassium hydroxide reacted is (molarity)×(volume) = 0.215×30.7 = 6.605 millimoles .

Moles of acetic acid = moles of potassium hydroxide .

Molarity of acetic acid = 6.605/29 = 0.227M

3 ans

From stoichiometric equation one mole of sulfuric acid reacts with 2 moles of sodium hydroxide . Moles of sodium hydroxide reacted = 0.156×42.8 = 6.676 millimoles . Moles of sulfuricacid reacted is = 6.676 millimoles = m×15.5 ...... m = 0.430 molar.

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