Question

Please solve q 127?

Answer 1

Please see the explanation below

Explanation:

The molecular mass of sulfuric acid is

`M_(H_2SO_4)=2+32+64=98g`

`"molality" ="moles solute"/"kg solvent"`

`90g` of `H_2SO_4` in `100 mL` solution

Number of moles is `=90/98=0.9184 "moles"`

Density of solution `=1.8 gmL^-1`

`180g (100ml)` of solution contains `=90g H_2SO_4` and `=90g H_2O`

molality `=0.9184/(90*10^-3)=10.2 "moles " kg ^-1`

Answer 2

As-written, the question has no correct answer choice. Choice `(c)` is the molarity, and not the molality.

`a)` `1.8`
`b)` `48.4`
`c)` `9.18`
`d)` `91.8`

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The original question was:

The molality of 90% weight/volume H2SO4 solution is (density = 1.8 g/mL)

And we treat it as-written.

`90%"w/v" = ("90 g H"_2"SO"_4)/("100 mL solution")`

The solution is solute plus solvent, and mass is additive. Therefore:

`100 cancel"mL soln" xx "1.8 g"/cancel"mL" = "180 g solution"`

`= "90 g solute" + "90 g solvent"`

We said there was `"90 g solute"` (here it does not matter, but we choose water as the solvent just because they are the same mass), so

`-> color(blue)("molality") = (90 cancel("g H"_2"SO"_4) xx "1 mol"/(98.079 cancel"g"))/(90 cancel"g solvent" xx "1 kg"/(1000 cancel"g"))`

`= "0.9176 mols solute"/"0.090 kg solvent"`

`=` `color(blue)("10.2 mol/kg")`

As a bonus, the molarity is...

`color(blue)("molarity") = "0.9176 mols solute"/"0.100 L soln"`

`=` `color(blue)("9.18 mols/L")`

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In addition, the molarity of `96%` `"H"_2"SO"_4` is `"18.01 M"`, because that is `%"w/w"`. I'll let you work that out for yourself though...