Calculate molarity of a 20% w/w aqueous solution of sulphuric acid?

Answer 1

#"2.04 mol/L"#

#%"w/w" = "Mass of solute"/"Mass of solution" × 100#
#20%"w/w"# means #"20 g"# of sulfuric acid is present in #"100 g"# of water
Density of water = #"1 g/mL"#
Volume of water in given solution = #"100 g"/"1 g/mL" = "100 mL" = color(blue)"0.1 L"#
Molar mass of #"H"_2"SO"_4 = "98 g/mol"#
Moles of #"H"_2"SO"_4# in given solution = #"20 g"/"98 g/mol" = color(blue)(10/49"mol")#
#"Molarity" = "Moles of solute"/"Volume of solution (in litres)"#
#"Molarity" = (10/49"mol")/"0.1 L" = color(red)"2.04 mol/L"#
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Answer 2
Relationship between #%"w/w"# and Molarity is
#"Molarity" = ("10" × %"w/w" × "density of solvent")/"Molar mass of solute"#
#"Molarity" = (10 × 20 × 1)/98 "mol/L" = "2.04 mol/L"#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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