An aqueous solution is 36.0 % by mass hydrochloric acid , HCl , and has a densit

Question

An aqueous solution is 36.0 % by mass hydrochloric acid, HCl, and has a density of 1.18 g/mL.

The mole fraction of hydrochloric acid in the solution is?

An aqueous solution of sodium acetate has a concentration of 8.04×10-2 molal.  

The percent by mass of sodium acetate in the solution is___ %.?

An aqueous solution of magnesium chloride, MgCl2, contains 1.36grams of magnesium chloride and 17.8 grams of water.

The percentage by mass of magnesium chloride in the solution is?___ %

An aqueous solution is made by dissolving 18.4 grams of iron(III) nitrate in 411 grams of water.  

The molality of iron(III) nitrate in the solution is? ___ m

Explanation / Answer

1.

36% by mass; it is (m/m) or (m/V)

I am calculating it as (m/m)

36% HCl solution means

36 g of HCl in 100 g of solution

So, mass of water = 100 g – 36 g = 64 g

Moles of HCl = (36 g) / (36.5 g/mol) = 0.99 mol
Moles of water = (64 g) / (18 g/mol) = 3.56 mol

Total moles = 0.99 mol + 3.56 mol = 4.55 mol

Moles fraction of HCl = (moles of HCl) / (Total moles)
                                       = 0.99 mol / 4.55 mol
                                       = 0.218

2.

[soldium acetate] = 8.04 x 10-2 molal = 0.0804 molal

So, in 1 Kg of solvent the moles of sodium acetate = 0.0804 moles

Or, in 1000 g of solvent the moles of sodium acetate = 0.0804 moles

Or, in 100 g of solvent the moles of sodium acetate = 0.00804 moles

Molar mass of sodium acetate = 82 g/mol

So, 1 mole of sodium acetate = 82 g

Or, 0.00804 moles of sodium acetate = 0.00804 x 82 g = 0.66 g

So, the percent by mass of sodium acetate in the solution is 0.66 %.

3.

Mass of MgCl2 = 1.36 g
Mass of water = 17.8 g

Total mass of solution = 1.36 g + 17.8 g = 19.16 g

So, in 19.16 g of solution mass of MgCl2 = 1.36 g
or, in 1 g of solution mass of MgCl2 = (1.36 / 19.16) g
or, in 100 g of solution mass of MgCl2 = (1.36 / 19.16) x 100 g
                                                                     = 7.10 g

So, percentage by mass of MgCl2 in the solution is 7.10 %

4.

Mass of iron(III) nitrate = 18.4 g

Molar mass of iron(III) nitrate = 241.86 g/mol

So, moles of iron(III) nitrate = (18.4 g) / (241.86 g/mol) = 0.076 mol

Now, in 411 g of water the moles of iron(III) nitrate = 0.076 mol
or, in 1 g of water the moles of iron(III) nitrate = (0.076 / 411) mol
or, in 1000 g of water the moles of iron(III) nitrate = (0.076 / 411) x 1000 mol
or, in 1 Kg of water the moles of iron(III) nitrate = 0.185 mol

So, the molality = 0.185 m

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