# Help with these three questions about mass and concentrations?

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#1)# If #"450 kg"# of #"SO"_2# reacts with coke (#"CS"_2# ) with #82%# consumption, what mass of coke is formed?

#2)# In a #"29 g"# mixture of #"BaO"# and #"CaO"# , #"100.8 mL"# of #"6 M HCl"# neutralizes the mixture. What is the #"%w/w BaO"# ?

#3)# If the true specific gravity of #"H"_2"SO"_4(l)# is #1.2# , then if we have a #27%"w/w"# solution of #"H"_2"SO"_4# , what is its molarity?

##### 2 Answers

Q-1

As per given condition

2 moles

or

or

Q-2

Let there be

The molar masses of

So by the condition

Again

So amount of HCl required is

Now by stoichiometry of the given reaction we see that number of moles of both

Hence

Multiplying [2] by 56 and subtracting the resulting equation from [1] we get

So mass of

So percentage of

Q-3

Given

So 100g solution contains 27g or

Or

Hence 1000mL solution contains

Hence molarity of the solution is

If the **true specific gravity** of the solution is *numerically the density of the solution*, i.e.

#"SG" = (rho_(H_2SO_4(aq)))/(rho_(H_2O(l))) = 1.2#

#=> rho_(H_2SO_4(aq)) = 1.2 cdot "1 g/mL" = "1.2 g/mL"#

Then, we are given:

#%"w/w" = ("270 g H"_2"SO"_4)/("1000 g soln")#

Since molarity is defined as

#"mols solute"/"L soln"# ,

we need the volume of the solution. Using the specific gravity of the solution, we have:

#V_(sol n) = 1000 cancel"g soln" xx cancel"1 mL"/(1.2 cancel"g soln") xx "1 L"/(1000 cancel"mL")#

#=# #"0.833 L soln"#

We next need the mols of solute:

#270 cancel("g H"_2"SO"_4) xx "1 mol"/(98.079 cancel("g H"_2"SO"_4))#

#= "2.753 mols H"_2"SO"_4#

This means we have a molarity of:

#color(blue)(["H"_2"SO"_4]) = ("2.753 mols H"_2"SO"_4)/("0.833 L soln")#

#=# #color(blue)("3.30 M")#

Either the answer key is incorrect or there's something off with my density. The rest is correct.